If polynomial Px - x2 - ax - b has factors x-a and x-b- where a- b - R- then the value of P2 is

X^2 + ax + b = (x a)(x b) ⇒ x^2 + ax + b =x^2 (a+b)x+ab ⇒(2a+b)x+b ab=0 ⇒ 2a + b = 0, or ab b=0 ⇒ 2a + b = 0, or b(a 1)=0 ⇒ 2a + b = 0, or b=0 , a=1 nbs ...

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Standard XI

Mathematics

Factor Theorem

If polynomial...

Question

If polynomial $P (x) = x^{2} + a x + b$ has factors $(x - a) and (x - b), where a, b \in R,$ then the value of $P (2)$ is

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Solution

The correct option is A $4$
$x^{2} + a x + b = (x - a) (x - b)$
$\Rightarrow x^{2} + a x + b = x^{2} - (a + b) x + a b$
$\Rightarrow (2 a + b) x + b - a b = 0$
$\Rightarrow 2 a + b = 0, or a b - b = 0$
$\Rightarrow 2 a + b = 0, or b (a - 1) = 0$
$\Rightarrow 2 a + b = 0, or b = 0, a = 1$
$\Rightarrow a = 1, b = - 2 or a = b = 0$
$\Rightarrow P (x) = x^{2} + x - 2 or P (x) = x^{2}$
$∴ P (2) = 4$

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If polynomial P(x)=x2+ax+b has factors (x−a) and (x−b), where a,b∈ R, then the value of P(2) is

The correct option is A 4x2+ax+b=(x−a)(x−b) ⇒x2+ax+b=x2−(a+b)x+ab ⇒(2a+b)x+b−ab=0 ⇒2a+b=0, or ab−b=0 ⇒2a+b=0, or b(a−1)=0 ⇒2a+b=0, or b=0,a=1 ⇒a=1,b=−2 or a=b=0 ⇒P(x)=x2+x−2 or P(x)=x2 ∴P(2)=4

If polynomial $P (x) = x^{2} + a x + b$ has factors $(x - a) and (x - b), where a, b \in R,$ then the value of $P (2)$ is