Question

If x2 – 3x + 2 is a factor of x^4-ax² + b then value of a, b one
(a)a=5,b = 4
(b) a = 5,6 = -4
(c) a = -5,6 = 4
(d) a=-5,b=4

Answer 1

Answer:

$\sf{(a) \ a=5, \ b=4}$

$\sf{Is \ the \ correct \ option.}$

Given:

• x²-3x+2 is a factor of p(x)x⁴-ax²+b

To find:

• The value of a and b.

Solution:

$\sf{Factorising \ polynomial \ x^{2}-3x+2}$

$\sf{\leadsto{x^{2}-3x+2}}$

$\sf{\leadsto{x^{2}-x-2x+2}}$

$\sf{\leadsto{x(x-1)-2(x-1)}}$

$\sf{\leadsto{(x-1)(x-2)}}$

$\sf{Hence, \ we \ can \ say \ (x-1) \ as \ well \ as}$

$\sf{(x-2) \ are \ factors \ of \ x^{4}-ax^{2}+b}$

$\sf{By \ Remainder \ theorem \ we \ get,}$

$\sf{p(1)=0 \ also \ p(2)=0}$

$\sf{p(1)=1^{4}-a(1)^{2}+b}$

$\sf{\therefore{1-a+b=0}}$

$\sf{\therefore{-a+b=-1...(1)}}$

$\sf{Also,}$

$\sf{p(2)=2^{4}-a(2)^{2}+b}$

$\sf{\therefore{16-4a+b=0}}$

$\sf{\therefore{-4a+b=-16...(2)}}$

$\sf{Subtracting \ equation (2) \ from \ equation (1)}$

$\sf{-a+b=-1}$

$\sf{-}$

$\sf{-4a+b=-16}$

_________________

$\sf{3a=15}$

$\sf{\therefore{a=\dfrac{15}{3}}}$

$\boxed{\sf{\therefore{a=5}}}$

$\sf{Substitute \ a=5 \ in \ equation (1) \ we \ get}$

$\sf{-5+b=-1}$

$\sf{\therefore{b=-1+5}}$

$\boxed{\sf{\therefore{b=4}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ a \ and \ b \ are}}}$

$\sf\purple{\tt{5 \ and \ 4 \ respectively.}}$