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 = 0, ab = b ⇒ nbsp;a = 1, b = 2 or a = b = 0 ⇒ P(x) = x^2 + x 2 or P(x) = x^2 ∴ P(2) = 4 ...

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Standard XI

Mathematics

Relation between Roots and Coefficients for Quadratic

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 D $4$
$x^{2} + a x + b = (x - a) (x - b)$
$x^{2} + a x + b = x^{2} - (a + b) x + a b$
$2 a + b = 0, a b = b \Rightarrow a = 1, b = - 2 or a = b = 0$
$\Rightarrow P (x) = x^{2} + x - 2 or P (x) = x^{2}$
$∴ P (2) = 4$

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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 D 4x2+ax+b=(x−a)(x−b) x2+ax+b=x2−(a+b)x+ab 2a+b=0,ab=b⇒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

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