| Exercise 2.1 Exercise 2.2 Exercise 2.3 |
Exercise 2.3
Question 1
Divide 2x3 + 3x + 1 by x + 2 and find the quotient and the reminder. Is q(x) a factor of 2x3 + 3x + 1 ?
Sol :Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Quotient = 2x2 – 4x + 11
Remainder = – 21
No, 2x2 – 4x + 11 is not a factor of 2x3 + 3x + 1 because remainder ≠ 0
Dividend = 3x3 + x2 + 2x + 5
Divisor = x2 + 2x + 1
Now, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Remainder = – 21
No, 2x2 – 4x + 11 is not a factor of 2x3 + 3x + 1 because remainder ≠ 0
Question 2
Divide 3x3 + x2 + 2x + 5 by 1 + 2x + x2 and find the quotient and the remainder. Is 1 + 2x + x2 a factor of 3x3 + x2 + 2x + 5
Sol :Dividend = 3x3 + x2 + 2x + 5
Divisor = x2 + 2x + 1
Now, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Quotient = 3x – 5
Remainder = 9x + 10
No, x2 + 2x + 1 is not a factor of 3x3 + x2 + 2x + 5 because remainder ≠ 0
Sol :
p(x) = x3 – 3x2 + 4x + 2,g(x) = x – 1
Dividend = x3 – 3x2 + 4x + 2
Divisor = x – 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Remainder = 9x + 10
No, x2 + 2x + 1 is not a factor of 3x3 + x2 + 2x + 5 because remainder ≠ 0
Question 3 A
Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :
p(x) = x3 – 3x2 + 4x + 2 , g(x) = x – 1Sol :
p(x) = x3 – 3x2 + 4x + 2,g(x) = x – 1
Dividend = x3 – 3x2 + 4x + 2
Divisor = x – 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Quotient [q(x)]= x2 – 2x + 2
Remainder [r(x)]= 4
Sol :
Sol :
p(x) = 2x4 + 3x3 + 4x2 + 19x + 45, g(x) = x – 2
Dividend = 2x4 + 3x3 + 4x2 + 19x + 45
Divisor = x – 2
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,
Remainder [r(x)]= 4
Question 3 B
Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :
p(x) = 4x3 – 3x2 + 2x + 3 , g(x) = x + 4Sol :
p(x) = 4x3 – 3x2 + 2x + 3,g(x) = x + 4
Dividend = 4x3 – 3x2 + 2x + 3
Divisor = x + 4
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Quotient = 4x2 – 13x + 54
Remainder = – 213
Dividend = 4x3 – 3x2 + 2x + 3
Divisor = x + 4
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Remainder = – 213
Question 3 C
Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :
p(x) = 2x4 + 3x3 + 4x2 + 19x + 45, g(x) = x – 2Sol :
p(x) = 2x4 + 3x3 + 4x2 + 19x + 45, g(x) = x – 2
Dividend = 2x4 + 3x3 + 4x2 + 19x + 45
Divisor = x – 2
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,
q(x) = 2x3+7x2+18x+55
r(x) = 155
Question 3 D
Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :
p(x) = x4 + 2x3 – 3x2 + x – 1, g(x) = x – 2Sol :
p(x) = x4 + 2x3 – 3x2 + x – 1, g(x) = x – 2
Dividend = x4 + 2x3 – 3x2 + x – 1
Divisor = x – 2
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,
q(x) = x3+4x2+5x+11
r(x) = 21
Sol :
p(x) = x3 – 3x2– x + 3, g(x) = x2 – 4x + 3
Dividend = x3 – 3x2– x + 3
Divisor = x2 – 4x + 3
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,
r(x) = 21
Question 3 E
Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :
p(x) = x3 – 3x2– x + 3, g(x) = x2 – 4x + 3Sol :
p(x) = x3 – 3x2– x + 3, g(x) = x2 – 4x + 3
Dividend = x3 – 3x2– x + 3
Divisor = x2 – 4x + 3
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,
q(x)= x + 1
r(x) = 0
Sol :
p(x) = x6 + x4 + x3 + x2 + 2x + 2, g(x) = x3 + 1
Dividend = x6 + x4 + x3 + x2 + 2x + 2
Divisor = x3 + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,
r(x) = 0
Question 3 F
Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :
p(x) = x6 + x4 + x3 + x2 + 2x + 2, g(x) = x3 + 1Sol :
p(x) = x6 + x4 + x3 + x2 + 2x + 2, g(x) = x3 + 1
Dividend = x6 + x4 + x3 + x2 + 2x + 2
Divisor = x3 + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,
q(x) = x3 + x
r(x) = x2 + x + 2
Sol :
p(x) = x6 + 3x2 + 10 , g(x) = x3 + 1
Dividend = x6 + 3x2 + 10
Divisor = x3 + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,
r(x) = x2 + x + 2
Question 3 G
Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :
p(x) = x6 + 3x2 + 10 and g(x) = x3 + 1Sol :
p(x) = x6 + 3x2 + 10 , g(x) = x3 + 1
Dividend = x6 + 3x2 + 10
Divisor = x3 + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,
q(x) = x3 – 1
r(x) = 3x2 + 11
Sol :
p(x) = x4 + 1, g(x) = x + 1
Dividend = x4 + 1,
Divisor = x + 1
r(x) = 3x2 + 11
Question 3 H
Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :
p(x) = x4 + 1, g(x) = x + 1Sol :
p(x) = x4 + 1, g(x) = x + 1
Dividend = x4 + 1,
Divisor = x + 1
q(x) = x3 – x2 + x – 1
r(x) = 0
On dividing x3 – 6x2 + 11x + k by x – 1 we get,
r(x) = 0
Question 4
By division process, find the value of k for which x – 1 is a factor of x3 – 6x2 + 11x + k.
Sol :On dividing x3 – 6x2 + 11x + k by x – 1 we get,
Since x – 1 is a factor of x3 – 6x2 + 11x + k,
This means x – 1 divides the given polynomial completely.
→ 6 x + k = 0
→ k = – 6x
Question 5
By division process, find the value of c for which 2x + 1 is a factor of 4x4 – 3x2 + 3x + c.
Sol :On dividing 4x4 – 3x2 + 3x + c by 2x + 1 we get,
Since 2x + 1 is a factor of 4x4 – 3x2 + 3x + c,
This means 2x + 1 divides the given polynomial completely,
→ 4x + c = o
→ c = – 4x
Sol :
p(x) = 2x2 + 3x + 1, g(x) = x + 2
Dividend = 2x2 + 3x + 1,
Divisor = x + 2
Apply the division algorithm we get,
This means 2x + 1 divides the given polynomial completely,
→ 4x + c = o
→ c = – 4x
Question 6 A
Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:
p(x) = 2x2 + 3x + 1, g(x) = x + 2Sol :
p(x) = 2x2 + 3x + 1, g(x) = x + 2
Dividend = 2x2 + 3x + 1,
Divisor = x + 2
Apply the division algorithm we get,
q(x) = 2x – 1
r(x) = 3
Sol :
p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Dividend = x3 – 3x2 + 5x – 3,
Divisor = x2 – 2
On applying division algorithm, we get,
q(x) = x – 3
r(x) = 7x – 9
Sol :
p(x) = x4 – 1,g(x) = x + 1
Dividend = x4 – 1
Divisor = x + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
r(x) = 3
Question 6 B
Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:
p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2Sol :
p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Dividend = x3 – 3x2 + 5x – 3,
Divisor = x2 – 2
On applying division algorithm, we get,
q(x) = x – 3
r(x) = 7x – 9
Question 6 C
Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:
p(x) = x4 – 1 , g(x) = x + 1Sol :
p(x) = x4 – 1,g(x) = x + 1
Dividend = x4 – 1
Divisor = x + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Quotient = x3 – x2 + x – 1
Remainder = 0
Sol :
p(x) = x3 – 3x2 + 4x + 2,g(x) = x – 1
Dividend = x3 – 3x2 + 4x + 2
Divisor = x – 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Remainder = 0
Question 6 D
Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:
p(x) = x3 – 3x2 + 4x + 2 , g(x) = x – 1Sol :
p(x) = x3 – 3x2 + 4x + 2,g(x) = x – 1
Dividend = x3 – 3x2 + 4x + 2
Divisor = x – 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Quotient = x – 5
Remainder = 13x + 7
Sol :
p(x) = x3 – 6x2 + 11x – 6,g(x) = x2 – 5x + 6
Dividend = x3 – 6x2 + 11x – 6
Divisor = x2 – 5x + 6
Here, dividend and divisor both are in the standard form.
Remainder = 13x + 7
Question 6 E
Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:
p(x) = x3 – 6x2 + 11x – 6 , g(x) = x2 – 5x + 6Sol :
p(x) = x3 – 6x2 + 11x – 6,g(x) = x2 – 5x + 6
Dividend = x3 – 6x2 + 11x – 6
Divisor = x2 – 5x + 6
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Quotient = x – 1
Remainder = 6x
Sol :
p(x) = 6x3 + 13x2 + x – 2,g(x) = 2x + 1
Dividend = 6x3 + 13x2 + x – 2
Divisor = 2x + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Question 6 F
Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:
p(x) = 6x3 + 13x2 + x – 2 , g(x) = 2x + 1Sol :
p(x) = 6x3 + 13x2 + x – 2,g(x) = 2x + 1
Dividend = 6x3 + 13x2 + x – 2
Divisor = 2x + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process
Quotient = 3x2 + 5x – 2
Remainder = 0
Sol :
Let us divide x3 + 3x2 – 12x + 4 by x – 2
The division process is
Remainder = 0
Question 7 A
Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:
x – 2, x3 + 3x3 – 12x + 4Sol :
Let us divide x3 + 3x2 – 12x + 4 by x – 2
The division process is
Here, the remainder is 0, therefore x – 2 is a factor of x3 + 3x2 – 12x + 4
Sol :
Let us divide 3x4 + 5x3 – 7x2 + 2x + 2 by x2 + 3x + 1
The division process is
Question 7 B
Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:
x2 + 3x + 1,3x4 + 5x3 – 7x2 + 2x + 2Sol :
Let us divide 3x4 + 5x3 – 7x2 + 2x + 2 by x2 + 3x + 1
The division process is
Here, the remainder is 0, therefore x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2
Question 7 C
Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:
x2 – 3x + 4,2x4 – 11x3 + 29x2 – 30x + 29Sol :
Let us divide 2x4 – 11x3 + 29x2 – 30x + 29 by x2 – 3x + 4
The division process is
Here, the remainder is 58x – 115 , therefore x2 – 3x + 4 is not a factor of 2x4 – 11x3 + 29x2 – 30x + 29
Sol :
Let us divide x3 – 3x2 – x + 3 by x2 – 4x + 3
The division process is
Question 7 D
Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:
x2 – 4x + 3,x3 – x3 – 3x4 – x + 3Sol :
Let us divide x3 – 3x2 – x + 3 by x2 – 4x + 3
The division process is
Here, the remainder is 0, therefore x2 – 4x + 3 is a factor of x3 – 3x2 – x + 3
Sol :
Let us divide t3 + t2 – 2t + 1 by t – 1
The division process is
Question 7 E
Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:
t – 1, t3 + t2 – 2t + 1Sol :
Let us divide t3 + t2 – 2t + 1 by t – 1
The division process is
Here, the remainder is 1, therefore t – 1 is not a factor of t3 + t2 – 2t + 1
Sol :
Let us divide t2 + 11t – 6 by t2 – 5t + 6
The division process is
Question 7 F
Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:
t2 – 5t + 6,t2 + 11t – 6Sol :
Let us divide t2 + 11t – 6 by t2 – 5t + 6
The division process is
Here, the remainder is 16t – 12,
Therefore, t2 – 5t + 6 is not a factor of t2 + 11t – 6
And also satisfying
(i) deg p(x) = deg q(x) + 1
(ii)deg q(x) = 1
(iii) deg q(x) = deg r(x) + 1
Sol :
(i)Let p(x) = 12x2 + 8x + 25, g(x) = 4,
q(x) = 3x2 + 2x + 6 , r(x) = 0
Here, degree p(x) = degree q(x) = 2
Now, g(x).q(x) + r(x) = (3x2 + 2x + 6)×4 + 1
= 12x2 + 8x + 24 + 1
= 12x2 + 8x + 25
(ii) Let p(x) = t3 + t2 – 2t, g(x) = t2 + 2t,
q(x) = t – 1 , r(x) = 0
Here, degree q(x) = 1
Now, g(x).q(x) + r(x) = (t2 + 2t)×(t – 1) + 0
= t3 – t2 + 2t2 – 2t
= t3 + t2 – 2t
(iii) Let p(x) = x3 + x2 + x + 1 , g(x) = x2 – 1,
q(x) = x + 1 , r(x) = 2x + 2
Here, degree q(x) = degree r(x) + 1 = 1
Now, g(x).q(x) + r(x) = (x2 – 1)×(x + 1) + 2x + 2
= x3 + x2 – x – 1 + 2x + 2
= x3 + x2 + x + 1
Sol :
Given zeroes is 3
So, (x – 3) is the factor of x3 – 6x2 + 11x – 6
Let us divide x3 – 6x2 + 11x – 6 by x – 3
The division process is
Therefore, t2 – 5t + 6 is not a factor of t2 + 11t – 6
Question 8
Give examples of polynomials p(x), g(x), q(x) and r(x) satisfying the Division Algorithm
p(x) = g(x).q(x) + r(x),deg r(x)<deg g(x)And also satisfying
(i) deg p(x) = deg q(x) + 1
(ii)deg q(x) = 1
(iii) deg q(x) = deg r(x) + 1
Sol :
(i)Let p(x) = 12x2 + 8x + 25, g(x) = 4,
q(x) = 3x2 + 2x + 6 , r(x) = 0
Here, degree p(x) = degree q(x) = 2
Now, g(x).q(x) + r(x) = (3x2 + 2x + 6)×4 + 1
= 12x2 + 8x + 24 + 1
= 12x2 + 8x + 25
(ii) Let p(x) = t3 + t2 – 2t, g(x) = t2 + 2t,
q(x) = t – 1 , r(x) = 0
Here, degree q(x) = 1
Now, g(x).q(x) + r(x) = (t2 + 2t)×(t – 1) + 0
= t3 – t2 + 2t2 – 2t
= t3 + t2 – 2t
(iii) Let p(x) = x3 + x2 + x + 1 , g(x) = x2 – 1,
q(x) = x + 1 , r(x) = 2x + 2
Here, degree q(x) = degree r(x) + 1 = 1
Now, g(x).q(x) + r(x) = (x2 – 1)×(x + 1) + 2x + 2
= x3 + x2 – x – 1 + 2x + 2
= x3 + x2 + x + 1
Question 9 A
Find all the zeroes of the polynomial given below having given numbers as its zeroes.
x3 – 6x2 + 11x – 6;3Sol :
Given zeroes is 3
So, (x – 3) is the factor of x3 – 6x2 + 11x – 6
Let us divide x3 – 6x2 + 11x – 6 by x – 3
The division process is
Here, quotient = x2 – 3x + 2
= x2 – 2x – x + 2
= x(x – 2) – 1(x – 2)
= (x – 1)(x – 2)
So, the zeroes are 1 and 2
Hence, all the zeroes of the given polynomial are 1, 2 and 3.
Sol :
Given zeroes are 1 and 2
So, (x – 1)and (x – 2) are the factors of x4 – 8x3 + 23x2 – 28x + 12
⟹ (x – 1)(x – 2) = x2 – 3x + 2 is a factor of given polynomial.
Consequently, x2 – 3x + 2 is also a factor of the given polynomial.
Now, let us divide x4 – 8x3 + 23x2 – 28x + 12 by x2 – 3x + 2
The division process is
= x2 – 2x – x + 2
= x(x – 2) – 1(x – 2)
= (x – 1)(x – 2)
So, the zeroes are 1 and 2
Hence, all the zeroes of the given polynomial are 1, 2 and 3.
Question 9 B
Find all the zeroes of the polynomial given below having given numbers as its zeroes.
x4 – 8x3 + 23x2 – 28x + 12;1,2Sol :
Given zeroes are 1 and 2
So, (x – 1)and (x – 2) are the factors of x4 – 8x3 + 23x2 – 28x + 12
⟹ (x – 1)(x – 2) = x2 – 3x + 2 is a factor of given polynomial.
Consequently, x2 – 3x + 2 is also a factor of the given polynomial.
Now, let us divide x4 – 8x3 + 23x2 – 28x + 12 by x2 – 3x + 2
The division process is
Here, quotient = x2 – 5x + 6
= x2 – 2x – 3x + 6
= x(x – 2) – 3(x – 2)
= (x – 3)(x – 2)
So, the zeroes are 3 and 2
Hence, all the zeroes of the given polynomial are 1, 2 ,2 and 3.
Sol :
Given zeroes is – 2
So, (x + 2) is the factor of x3 + 2x2 – x – 2
Let us divide x3 + 2x2 – x – 2 by x + 2
The division process is
= x2 – 2x – 3x + 6
= x(x – 2) – 3(x – 2)
= (x – 3)(x – 2)
So, the zeroes are 3 and 2
Hence, all the zeroes of the given polynomial are 1, 2 ,2 and 3.
Question 9 C
Find all the zeroes of the polynomial given below having given numbers as its zeroes.
x3 + 2x2 – 2; – 2Sol :
Given zeroes is – 2
So, (x + 2) is the factor of x3 + 2x2 – x – 2
Let us divide x3 + 2x2 – x – 2 by x + 2
The division process is
Here, quotient = x2 – 1
= (x – 1)(x + 1)
So, the zeroes are – 1 and 1
Hence, all the zeroes of the given polynomial are – 1, – 2 and 1.
Sol :
Given zeroes is – 3
So, (x + 3) is the factor of x3 + 5x2 + 7x + 3
Let us divide x3 + 5x2 + 7x + 3 by x + 3
The division process is
= (x – 1)(x + 1)
So, the zeroes are – 1 and 1
Hence, all the zeroes of the given polynomial are – 1, – 2 and 1.
Question 9 D
Find all the zeroes of the polynomial given below having given numbers as its zeroes.
x3 + 5x2 + 7x + 3; – 3Sol :
Given zeroes is – 3
So, (x + 3) is the factor of x3 + 5x2 + 7x + 3
Let us divide x3 + 5x2 + 7x + 3 by x + 3
The division process is
Here, quotient = x2 + 2x + 1
= (x + 1)2
So, the zeroes are – 1 and – 1
Hence, all the zeroes of the given polynomial are – 1, – 1 and – 3.
Sol :
x4 – 6x3 – 26x2 + 138x – 35;2±√3
Given zeroes are 2 + √3 and 2 – √3
So, (x – 2 – √3)and (x – 2 + √3) are the factors of x4 – 6x3 – 26x2 + 138x – 35
⟹ (x – 2 – √3)(x – 2 + √3)
= x2 – 2x + √3 x – 2x + 4 – 2√3 – √3 x + 2√3 – 3
= x2 – 4x + 1 is a factor of given polynomial.
Consequently, x2 – 4x + 1 is also a factor of the given polynomial.
Now, let us divide x4 – 6x3 – 26x2 + 138x – 35 by x2 – 4x + 1
= (x + 1)2
So, the zeroes are – 1 and – 1
Hence, all the zeroes of the given polynomial are – 1, – 1 and – 3.
Question 9 E
Find all the zeroes of the polynomial given below having given numbers as its zeroes.
x4 – 6x3 – 26x2 + 138x – 35;2±√3Sol :
x4 – 6x3 – 26x2 + 138x – 35;2±√3
Given zeroes are 2 + √3 and 2 – √3
So, (x – 2 – √3)and (x – 2 + √3) are the factors of x4 – 6x3 – 26x2 + 138x – 35
⟹ (x – 2 – √3)(x – 2 + √3)
= x2 – 2x + √3 x – 2x + 4 – 2√3 – √3 x + 2√3 – 3
= x2 – 4x + 1 is a factor of given polynomial.
Consequently, x2 – 4x + 1 is also a factor of the given polynomial.
Now, let us divide x4 – 6x3 – 26x2 + 138x – 35 by x2 – 4x + 1
The division process is
Here, quotient = x2 – 2x – 35
= x2 – 7x + 5x – 35
= x(x – 7) + 5(x – 7)
= (x + 5)(x – 7)
So, the zeroes are – 5 and 7
Hence, all the zeroes of the given polynomial are – 5, 7, 2+√3 and 2-√3
Sol :
x4 + x3 – 34x2 – 4x + 120;2, – 2.
Given zeroes are – 2 and 2
So, (x + 2)and (x – 2) are the factors of x4 + x3 – 34x2 – 4x + 120
⟹ (x + 2)(x – 2) = x2 – 4 is a factor of given polynomial.
Consequently, x2 – 4 is also a factor of the given polynomial.
Now, let us divide x4 + x3 – 34x2 – 4x + 120 by x2 – 4
The division process is
= x2 – 7x + 5x – 35
= x(x – 7) + 5(x – 7)
= (x + 5)(x – 7)
So, the zeroes are – 5 and 7
Hence, all the zeroes of the given polynomial are – 5, 7, 2+√3 and 2-√3
Question 9 F
Find all the zeroes of the polynomial given below having given numbers as its zeroes.
x4 + x3 – 34x2 – 4x + 120;2, – 2.Sol :
x4 + x3 – 34x2 – 4x + 120;2, – 2.
Given zeroes are – 2 and 2
So, (x + 2)and (x – 2) are the factors of x4 + x3 – 34x2 – 4x + 120
⟹ (x + 2)(x – 2) = x2 – 4 is a factor of given polynomial.
Consequently, x2 – 4 is also a factor of the given polynomial.
Now, let us divide x4 + x3 – 34x2 – 4x + 120 by x2 – 4
The division process is
Here, quotient =x2 + x – 30
= x2 + 6x – 5x – 30
= x(x + 6) – 5(x + 6)
= (x + 6)(x – 5)
So, the zeroes are – 6 and 5
Hence, all the zeroes of the given polynomial are – 2 , – 6, 2 and 5.
Sol :
2x4 + 7x3 – 19x2 – 14x + 30;√(2, – √2)
Given zeroes are √2 and – √2
So, (x – √2)and (x + √2) are the factors of 2x4 + 7x3 – 19x2 – 14x + 30
⟹ (x – √2)(x + √2) = x2 – 2 is a factor of given polynomial.
Consequently, x2 – 2 is also a factor of the given polynomial.
Now, let us divide 2x4 + 7x3 – 19x2 – 14x + 30 by x2 – 2
The division process is
= x2 + 6x – 5x – 30
= x(x + 6) – 5(x + 6)
= (x + 6)(x – 5)
So, the zeroes are – 6 and 5
Hence, all the zeroes of the given polynomial are – 2 , – 6, 2 and 5.
Question 9 G
Find all the zeroes of the polynomial given below having given numbers as its zeroes.
2x4 + 7x3 – 19x2 – 14x + 30;√2, – √2Sol :
2x4 + 7x3 – 19x2 – 14x + 30;√(2, – √2)
Given zeroes are √2 and – √2
So, (x – √2)and (x + √2) are the factors of 2x4 + 7x3 – 19x2 – 14x + 30
⟹ (x – √2)(x + √2) = x2 – 2 is a factor of given polynomial.
Consequently, x2 – 2 is also a factor of the given polynomial.
Now, let us divide 2x4 + 7x3 – 19x2 – 14x + 30 by x2 – 2
The division process is
Here, quotient =2x2 + 7x – 15
= 2x2 + 10x – 3x – 15
= 2x(x + 5) – 3(x + 5)
= (2x – 3)(x + 5)
So, the zeroes are – 5 and $\frac{3}{2}$
Hence, all the zeroes of the given polynomial are – 5, – √2, √2 and $\frac{3}{2}$
Sol :
2x4 – 9x3 + 5x2 + 3x – 1;2±√3
Given zeroes are 2 + √3 and 2 – √3
So, (x – 2 – √3)and (x – 2 + √3) are the factors of 2x4 – 9x3 + 5x2 + 3x – 1
⟹ (x – 2 – √3)(x – 2 + √3)
= x2 – 2x + √3 x – 2x + 4 – 2√3 – √3 x + 2√3 – 3
= x2 – 4x + 1 is a factor of given polynomial.
Consequently, x2 – 4x + 1 is also a factor of the given polynomial.
Now, let us divide 2x4 – 9x3 + 5x2 + 3x – 1 by x2 – 4x + 1
The division process is
= 2x2 + 10x – 3x – 15
= 2x(x + 5) – 3(x + 5)
= (2x – 3)(x + 5)
So, the zeroes are – 5 and $\frac{3}{2}$
Hence, all the zeroes of the given polynomial are – 5, – √2, √2 and $\frac{3}{2}$
Question 9 H
Find all the zeroes of the polynomial given below having given numbers as its zeroes.
2x4 – 9x3 + 5x2 + 3x – 1;2±√3Sol :
2x4 – 9x3 + 5x2 + 3x – 1;2±√3
Given zeroes are 2 + √3 and 2 – √3
So, (x – 2 – √3)and (x – 2 + √3) are the factors of 2x4 – 9x3 + 5x2 + 3x – 1
⟹ (x – 2 – √3)(x – 2 + √3)
= x2 – 2x + √3 x – 2x + 4 – 2√3 – √3 x + 2√3 – 3
= x2 – 4x + 1 is a factor of given polynomial.
Consequently, x2 – 4x + 1 is also a factor of the given polynomial.
Now, let us divide 2x4 – 9x3 + 5x2 + 3x – 1 by x2 – 4x + 1
The division process is
Here, quotient = 2x2 – x – 1
= 2x2 – 2x + x – 1
= 2x(x – 1) + 1(x – 1)
= (2x + 1)(x – 1)
So, the zeroes are $-\frac{1}{2}$ and 1
Hence, all the zeroes of the given polynomial are $-\frac{1}{2}$, 1, 2 + √3 and 2 – √3
Sol :
2x3 – 4x – x2 + 2;√2, – √2
Given zeroes are √2 and – √2
So, (x – √2) and (x + √2) are the factors of 2x3 – 4x – x2 + 2
⟹ (x – √2)(x + √2) = x2 – 2 is a factor of given polynomial.
Consequently, x2 – 2 is also a factor of the given polynomial.
Now, let us divide 2x3 – 4x – x2 + 2by x2 – 2
The division process is
= 2x2 – 2x + x – 1
= 2x(x – 1) + 1(x – 1)
= (2x + 1)(x – 1)
So, the zeroes are $-\frac{1}{2}$ and 1
Hence, all the zeroes of the given polynomial are $-\frac{1}{2}$, 1, 2 + √3 and 2 – √3
Question 9 I
Find all the zeroes of the polynomial given below having given numbers as its zeroes.
2x3 – 4x – x2 + 2;√2, – √22x3 – 4x – x2 + 2;√2, – √2
Given zeroes are √2 and – √2
So, (x – √2) and (x + √2) are the factors of 2x3 – 4x – x2 + 2
⟹ (x – √2)(x + √2) = x2 – 2 is a factor of given polynomial.
Consequently, x2 – 2 is also a factor of the given polynomial.
Now, let us divide 2x3 – 4x – x2 + 2by x2 – 2
The division process is
Here, quotient = 2x – 1
So, the zeroes is $\frac{1}{2}$
Hence, all the zeroes of the given polynomial are -√2,√2 and $\frac{1}{2}$
Hence, all the zeroes of the given polynomial are -√2,√2 and $\frac{1}{2}$
Question 10
Verify that 3,-1, $-\frac{1}{3}$ are the zeroes of the cubic polynomial p(x) = 3x2 – 5x2 – 11x – 3 and then verify the relationship between the zeroes and the coefficients.
Sol :Let p(x) = 3 x3 – 5 x2 – 11x – 3
Then, p( – 1) = 3( – 1)3 – 5( – 1)2 – 11( – 1) – 3
= – 3 – 5 + 11 – 3
= 0
$p\left(-\frac{1}{3}\right)=3\left(-\frac{1}{3}\right)^{3}-5\left(-\frac{1}{3}\right)^{2}-11\left(-\frac{1}{3}\right)-3$
$=\left(-\frac{1}{9}\right)-\left(\frac{5}{9}\right)+\left(\frac{11}{3}\right)-3$
$=\left(\frac{-1-5+33-27}{9}\right)$
= 0
p(3) = 3(3)3 – 5(3)2 – 11(3) – 3
= 81 – 45 – 33 – 3
= 0
Hence, we verified that 3, – 1 and $-\frac{1}{3}$ are the zeroes of the given polynomial.
So, we take α = 3, β = – 1, $\gamma=-\frac{1}{3}$
Verification
α + β + γ $=3+(-1)+\left(-\frac{1}{3}\right)=\left(\frac{5}{3}\right)$
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{5}{3}$
αβ + βγ + γα $=(3)(-1)+(-1)\left(-\frac{1}{3}\right)+\left(-\frac{1}{3}\right)(3)$
$=\left(-\frac{11}{3}\right)$
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{-11}{3}=\left(-\frac{11}{3}\right)$
and αβγ $=3 \times-1 \times\left(-\frac{1}{3}\right)$
= 1
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-3}{3}=1$
Thus, the relationship between the zeroes and the coefficients is verified.
Sol :
Let p(x) = x3 – 4x2 + 5x – 2
Then, p(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2
= 0
p(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
= 0
Hence, 2, 1 and 1 are the zeroes of the given polynomial x3 – 4x2 + 5x – 2.
Now, Let α = 2 , β = 1 and γ = 1
Then, α + β + γ = 2 + 1 + 1 = 4
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-4}{1}=4$
αβ + βγ + γα = (2)(1) + (1)(1) + (1)(2)
= 2 + 1 + 2
= 5
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{5}{1}=5$
and αβγ = 2 × 1 × 1
= 2
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-2}{1}=2$
Thus, the relationship between the zeroes and the coefficients is verified.
Sol :
Let p(x) = x3 – 6x2 + 11x – 6
Then, p(1) = (1)3 – 6(1)2 + 11(1) – 6
= 1 – 6 + 11 – 6
= 0
p(2) = (2)3 – 6(2)2 + 11(2) – 6
= 8 – 24 + 22 – 6
= 0
p(3) = (3)3 – 6(3)2 + 11(3) – 6
= 27 – 54 + 33 – 6
= 0
Hence, 1, 2 and 3 are the zeroes of the given polynomial x3 – 6x2 + 11x – 6.
Now, Let α = 1 , β = 2 and γ = 3
Then, α + β + γ = 1 + 2 + 3 = 6
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-6}{1}=6$
αβ + βγ + γα = (1)(2) + (2)(3) + (3)(1)
= 2 + 6 + 3
= 11
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{11}{1}=11$
and αβγ = 1 × 2 × 3
= 6
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-6}{1}=6$
Thus, the relationship between the zeroes and the coefficients is verified.
Sol :
Let p(x) = x3 + 2x2 – x – 2
Then, p( – 2) = ( – 2)3 + 2( – 2)2 – ( – 2) – 2
= – 8 + 8 + 2 – 2
= 0
p(1) = (1)3 + 2(1)2 – (1) – 2
= 1 + 2 – 1 – 2
= 0
Hence, – 2, – 2 and 1 are the zeroes of the given polynomial x3 + 2x2 – x – 2.
Now, Let α = – 2 , β = – 2 and γ = 1
Then, α + β + γ = – 2 + ( – 2) + 1 = – 3
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{2}{1}=-2$
αβ + βγ + γα = ( – 2)( – 2) + ( – 2)(1) + (1)( – 2)
= 4 – 2 – 2
= 0
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{-1}{1}=-1$
and αβγ = ( – 2) × ( – 2) × 1
= 4
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-2}{1}=2$
Thus, the relationship between the zeroes and the coefficients is not verified.
Sol :
Let p(x) = x3 + 5x2 + 7x + 3.
Then, p( – 1) = ( – 1)3 + 5( – 1)2 + 7( – 1) + 3
= – 1 + 5 – 7 + 3
= 0
p( – 3) = ( – 3)3 + 5( – 3)2 + 7( – 3) + 3
= – 27 + 45 – 21 + 3
= 0
Hence, – 1, – 1 and – 3 are the zeroes of the given polynomial x3 + 5x2 + 7x + 3.
Now, Let α = – 1 , β = – 1 and γ = – 3
Then, α + β + γ = – 1 + ( – 1) + ( – 3) = – 5
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{5}{1}=-5$
αβ + βγ + γα = ( – 1)( – 1) + ( – 1)( – 3) + ( – 3)( – 1)
= 1 + 3 + 3
= 7
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{7}{1}=7$
and αβγ = ( – 1) × ( – 1) × ( – 3)
= – 3
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{3}{1}=-3$
Thus, the relationship between the zeroes and the coefficients is verified.
Let the zeroes of the cubic polynomial be
α = 1, β = 2 and γ = 3
Then, α + β + γ = 1 + 2 + 3 = 6
αβ + βγ + γα = (1)(2) + (2)(3) + (3)(1)
= 2 + 6 + 3
= 11
and αβγ = 1 × 2 × 3
= 6
Now, required cubic polynomial
= x3 – (α + β + γ) x2 + (αβ + βγ + γα)x – αβγ
= x3 – (6) x2 + (11)x – 6
= x3 – 6 x2 + 11x – 6
So, x3 – 6 x2 + 11x – 6 is the required cubic polynomial which satisfy the given conditions.
Let the zeroes of the cubic polynomial be
α = – 3, β = – 2 and γ = 2
Then, α + β + γ = – 3 + ( – 2) + 2
= – 3 – 2 + 2
= – 3
αβ + βγ + γα = ( – 3)( – 2) + ( – 2)(2) + (2)( – 3)
= 6 – 4 – 6
= – 4
and αβγ = ( – 3) × ( – 2) × 2
= 6 × 2
= 12
Now, required cubic polynomial
= x3 – (α + β + γ) x2 + (αβ + βγ + γα)x – αβγ
= x3 – ( – 3) x2 + ( – 4)x – 12
= x3 + 3 x2 – 4x – 12
So, x3 + 3x^2 – 4x – 12 is the required cubic polynomial which satisfy the given conditions.
Let the zeroes be α, β and γ.
Then, we have
α + β + γ = 0
αβ + βγ + γα = – 7
and αβγ = – 6
Now, required cubic polynomial
= x3 – (α + β + γ) x2 + (αβ + βγ + γα)x – αβγ
= x3 – (0) x2 + ( – 7)x – ( – 6)
= x3 – 7x + 6
So, x3 – 7x + 6 is the required cubic polynomial.
Sol :
Let the zeroes be α, β and γ.
Then, we have
α + β + γ = 2
αβ + βγ + γα = – 7
and αβγ = – 14
αβ + βγ + γα $=(3)(-1)+(-1)\left(-\frac{1}{3}\right)+\left(-\frac{1}{3}\right)(3)$
$=\left(-\frac{11}{3}\right)$
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{-11}{3}=\left(-\frac{11}{3}\right)$
and αβγ $=3 \times-1 \times\left(-\frac{1}{3}\right)$
= 1
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-3}{3}=1$
Thus, the relationship between the zeroes and the coefficients is verified.
Question 11 A
Verify that the numbers given alongside of the cubic polynomial are their zeros. Also verify the relationship between the zeroes and the coefficients in each case :
x3 – 4x2 + 5x – 2;2, 1, 1Let p(x) = x3 – 4x2 + 5x – 2
Then, p(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2
= 0
p(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
= 0
Hence, 2, 1 and 1 are the zeroes of the given polynomial x3 – 4x2 + 5x – 2.
Now, Let α = 2 , β = 1 and γ = 1
Then, α + β + γ = 2 + 1 + 1 = 4
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-4}{1}=4$
αβ + βγ + γα = (2)(1) + (1)(1) + (1)(2)
= 2 + 1 + 2
= 5
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{5}{1}=5$
and αβγ = 2 × 1 × 1
= 2
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-2}{1}=2$
Thus, the relationship between the zeroes and the coefficients is verified.
Question 11 B
Verify that the numbers given alongside of the cubic polynomial are their zeros. Also verify the relationship between the zeroes and the coefficients in each case :
x3 – 6x2 + 11x – 6 ;1, 2, 3Sol :
Let p(x) = x3 – 6x2 + 11x – 6
Then, p(1) = (1)3 – 6(1)2 + 11(1) – 6
= 1 – 6 + 11 – 6
= 0
p(2) = (2)3 – 6(2)2 + 11(2) – 6
= 8 – 24 + 22 – 6
= 0
p(3) = (3)3 – 6(3)2 + 11(3) – 6
= 27 – 54 + 33 – 6
= 0
Hence, 1, 2 and 3 are the zeroes of the given polynomial x3 – 6x2 + 11x – 6.
Now, Let α = 1 , β = 2 and γ = 3
Then, α + β + γ = 1 + 2 + 3 = 6
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-6}{1}=6$
αβ + βγ + γα = (1)(2) + (2)(3) + (3)(1)
= 2 + 6 + 3
= 11
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{11}{1}=11$
and αβγ = 1 × 2 × 3
= 6
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-6}{1}=6$
Thus, the relationship between the zeroes and the coefficients is verified.
Question 11 C
Verify that the numbers given alongside of the cubic polynomial are their zeros. Also verify the relationship between the zeroes and the coefficients in each case :
x3 + 2x2 – x – 2; – 2 – 2, 1Sol :
Let p(x) = x3 + 2x2 – x – 2
Then, p( – 2) = ( – 2)3 + 2( – 2)2 – ( – 2) – 2
= – 8 + 8 + 2 – 2
= 0
p(1) = (1)3 + 2(1)2 – (1) – 2
= 1 + 2 – 1 – 2
= 0
Hence, – 2, – 2 and 1 are the zeroes of the given polynomial x3 + 2x2 – x – 2.
Now, Let α = – 2 , β = – 2 and γ = 1
Then, α + β + γ = – 2 + ( – 2) + 1 = – 3
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{2}{1}=-2$
αβ + βγ + γα = ( – 2)( – 2) + ( – 2)(1) + (1)( – 2)
= 4 – 2 – 2
= 0
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{-1}{1}=-1$
and αβγ = ( – 2) × ( – 2) × 1
= 4
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-2}{1}=2$
Thus, the relationship between the zeroes and the coefficients is not verified.
Question 11 D
Verify that the numbers given alongside of the cubic polynomial are their zeros. Also verify the relationship between the zeroes and the coefficients in each case :
x3 + 5x2 + 7x + 3 ; – 3, 2 – 1, – 1Sol :
Let p(x) = x3 + 5x2 + 7x + 3.
Then, p( – 1) = ( – 1)3 + 5( – 1)2 + 7( – 1) + 3
= – 1 + 5 – 7 + 3
= 0
p( – 3) = ( – 3)3 + 5( – 3)2 + 7( – 3) + 3
= – 27 + 45 – 21 + 3
= 0
Hence, – 1, – 1 and – 3 are the zeroes of the given polynomial x3 + 5x2 + 7x + 3.
Now, Let α = – 1 , β = – 1 and γ = – 3
Then, α + β + γ = – 1 + ( – 1) + ( – 3) = – 5
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{5}{1}=-5$
αβ + βγ + γα = ( – 1)( – 1) + ( – 1)( – 3) + ( – 3)( – 1)
= 1 + 3 + 3
= 7
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{7}{1}=7$
and αβγ = ( – 1) × ( – 1) × ( – 3)
= – 3
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{3}{1}=-3$
Thus, the relationship between the zeroes and the coefficients is verified.
Question 12
Find a cubic polynomial having 1, 2, 3 as its zeroes.
Sol :Let the zeroes of the cubic polynomial be
α = 1, β = 2 and γ = 3
Then, α + β + γ = 1 + 2 + 3 = 6
αβ + βγ + γα = (1)(2) + (2)(3) + (3)(1)
= 2 + 6 + 3
= 11
and αβγ = 1 × 2 × 3
= 6
Now, required cubic polynomial
= x3 – (α + β + γ) x2 + (αβ + βγ + γα)x – αβγ
= x3 – (6) x2 + (11)x – 6
= x3 – 6 x2 + 11x – 6
So, x3 – 6 x2 + 11x – 6 is the required cubic polynomial which satisfy the given conditions.
Question 13
Find a cubic polynomial having – 3, – 2, 2 as its zeroes.
Sol :Let the zeroes of the cubic polynomial be
α = – 3, β = – 2 and γ = 2
Then, α + β + γ = – 3 + ( – 2) + 2
= – 3 – 2 + 2
= – 3
αβ + βγ + γα = ( – 3)( – 2) + ( – 2)(2) + (2)( – 3)
= 6 – 4 – 6
= – 4
and αβγ = ( – 3) × ( – 2) × 2
= 6 × 2
= 12
Now, required cubic polynomial
= x3 – (α + β + γ) x2 + (αβ + βγ + γα)x – αβγ
= x3 – ( – 3) x2 + ( – 4)x – 12
= x3 + 3 x2 – 4x – 12
So, x3 + 3x^2 – 4x – 12 is the required cubic polynomial which satisfy the given conditions.
Question 14
Find a cubic polynomial with the sum of its zeroes are 0, – 7 and – 6 respectively.
Sol :Let the zeroes be α, β and γ.
Then, we have
α + β + γ = 0
αβ + βγ + γα = – 7
and αβγ = – 6
Now, required cubic polynomial
= x3 – (α + β + γ) x2 + (αβ + βγ + γα)x – αβγ
= x3 – (0) x2 + ( – 7)x – ( – 6)
= x3 – 7x + 6
So, x3 – 7x + 6 is the required cubic polynomial.
Question 15 A
Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:
2, – 7, – 14Sol :
Let the zeroes be α, β and γ.
Then, we have
α + β + γ = 2
αβ + βγ + γα = – 7
and αβγ = – 14
Now, required cubic polynomial
=x3-(α+β+γ)x2+(αβ+βγ+γα)-αβγ
=x3-(2)x2+(-7x)-(-14)
=x3-2x2-7x+14)
So
x3 – 2x2 – 7x + 14 is the required cubic polynomial.
Sol :
Let the zeroes be α, β and γ.
Then, we have
$\alpha+\beta+\gamma=-4$
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{2}$
$\alpha \beta \gamma=\frac{1}{3}$
Now, required cubic polynomial
$=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$
$=x^{3}-(-4) x^{2}+\left(\frac{1}{2}\right) x-\left(\frac{1}{3}\right)$
$=\frac{6 x^{3}+24 x^{2}+3 x-2}{6}$
So, 6x3 + 24x2 + 3x – 2 is the required cubic polynomial which satisfy the given conditions.
Sol :
Let the zeroes be α, β and γ.
Then, we have
$\alpha+\beta+\gamma=\frac{5}{7}$
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{7}$
$\alpha \beta \gamma=\frac{1}{7}$
x3 – 2x2 – 7x + 14 is the required cubic polynomial.Question 15 B
Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:
$-4 \frac{1}{2}, \frac{1}{3}$Sol :
Let the zeroes be α, β and γ.
Then, we have
$\alpha+\beta+\gamma=-4$
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{2}$
$\alpha \beta \gamma=\frac{1}{3}$
Now, required cubic polynomial
$=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$
$=x^{3}-(-4) x^{2}+\left(\frac{1}{2}\right) x-\left(\frac{1}{3}\right)$
$=\frac{6 x^{3}+24 x^{2}+3 x-2}{6}$
So, 6x3 + 24x2 + 3x – 2 is the required cubic polynomial which satisfy the given conditions.
Question 15 C
Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:
$\frac{5}{7}, \frac{1}{7}, \frac{1}{7}$Let the zeroes be α, β and γ.
Then, we have
$\alpha+\beta+\gamma=\frac{5}{7}$
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{7}$
$\alpha \beta \gamma=\frac{1}{7}$
Now, required cubic polynomial
$=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$
$=x^{3}-\left(\frac{5}{7}\right) x^{2}+\left(\frac{1}{7}\right) x-\left(\frac{1}{7}\right)$
$=\frac{7 x^{3}-5 x^{2}+x-1}{7}$
So, 7x3 – 5x2 + x – 1 is the required cubic polynomial which satisfy the given conditions.
Sol :
Let the zeroes be α, β and γ.
Then, we have
$\alpha+\beta+\gamma=\frac{2}{5}$
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{10}$
$\alpha \beta \gamma=\frac{1}{2}$
Now, required cubic polynomial
$=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$
$=x^{3}-\left(\frac{2}{5}\right) x^{2}+\left(\frac{1}{10}\right) x-\left(\frac{1}{2}\right)$
$=\frac{10 x^{3}-4 x^{2}+x-5}{10}$
So, 10x3 – 4x2 + x – 5 is the required cubic polynomial which satisfy the given conditions.
$=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$
$=x^{3}-\left(\frac{5}{7}\right) x^{2}+\left(\frac{1}{7}\right) x-\left(\frac{1}{7}\right)$
$=\frac{7 x^{3}-5 x^{2}+x-1}{7}$
So, 7x3 – 5x2 + x – 1 is the required cubic polynomial which satisfy the given conditions.
Question 15 D
Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:
$\frac{2}{5}, \frac{1}{10}, \frac{1}{2}$Let the zeroes be α, β and γ.
Then, we have
$\alpha+\beta+\gamma=\frac{2}{5}$
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{10}$
$\alpha \beta \gamma=\frac{1}{2}$
Now, required cubic polynomial
$=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$
$=x^{3}-\left(\frac{2}{5}\right) x^{2}+\left(\frac{1}{10}\right) x-\left(\frac{1}{2}\right)$
$=\frac{10 x^{3}-4 x^{2}+x-5}{10}$
So, 10x3 – 4x2 + x – 5 is the required cubic polynomial which satisfy the given conditions.
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
It's a good but some questions answers are wrong please crack this
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