"Question 2 Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x^3 + x^2 − 2x − 1, g(x) = x + 1 (ii) p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2 (iii) p(x) = x^3 − 4x^2 + x + 6, g(x) = x − 3
Class 9 - Math - Polynomials Page 43"
Answers
If (x-a) is a factor of p(x) then p(a)= 0
If (x+a) is a factor of p(x) then p(-a)= 0
If (ax-b) is a factor of p(x) then p(b/a)= 0
If (x-a) (x-b) is a factor of p(x) then p(a)= 0 & p(b) = 0.
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Solution:
If (x+1) (x+2) (x-3)is a factor of given polynomial say p(X) , then at x= -1, x= -2, x= 3, p(x) will become zero, otherwise it is not a factor of given polynomial.
i) If g(x) = x + 1 is a factor of given polynomial p(x),
p(-
1) must be zero.
p(x) = 2x³+x² -2x –
1
p(- 1) = 2(-1)3 + (-1)2 – 2(-1) – 1
=
2(- 1) + 1 + 2 – 1 = 0
Hence, g(x)= x + 1 is a factor of given polynomial.
(ii) If g(x) = x + 2 is a factor of given polynomial p(x), p(- 2) must be 0.
p(x) =x³+3x²+ 3x +
1
p(-2) =(-2)³+3(-2)²+3(-2)
+ 1
= -8 + 12 – 6 + 1
= -1
As, p(-2) ≠ 0
Hence g(x) = x + 2 is not a factor of given polynomial.
(iii) If g(x) = x – 3 is a factor of given
polynomial p(x), p(3) must be 0.
p(x) = x3 – 4x2 + x + 6
p(3) = (3)3 – 4(3)2 + 3 + 6
= 27 – 36 + 9 = 0
Hence, g(x) = x – 3 is a factor of given
polynomial.
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