Question

find the values of a and b so that ( x+1 ) and (x-1) are factors of
${x}^{4} \: + a {x}^{3} \: - 3 {x}^{2} \: + 2x \: + b \:$
class - 9

Answer 1

x) = x5 – x4 – 4x3 + 3x2 + 3x + b

q(x) = x3 + 2x2 + a 

Zeroes of q(x) are also the zeroes of p(x)

Þ q(x) is a factor of p(x)

To find the another factor of p(x), it should be divided by q(x).


x2 - 3x + 2 = (x – 2)(x – 1)

Therefore, 2 and 1 are the other zeroes of p(x).

Similarly, when p(x) is divided by x2 - 3x + 2, the quotient is x3 + 2x2 – 1 and the

Remainder is (b + 2).

x3 + 2x2 – 1 = x3 + 2x2 + a

⇒ a = -1

Remainder = 0

⇒ (b + 2) = 0

⇒ b = -2