If polynomial P(x) = x^2 + ax + b has factors | vedclass

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(D) Given the polynomial $P(x) = x^2 + ax + b$ has factors $(x - a)(x - b)$.Expanding the factors: $(x - a)(x - b) = x^2 - (a + b)x + ab$.Comparing th

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$4$

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(D) Given the polynomial $P(x) = x^2 + ax + b$ has factors $(x - a)(x - b)$.Expanding the factors: $(x - a)(x - b) = x^2 - (a + b)x + ab$.Comparing this with $P(x) = x^2 + ax + b$,we get:$a = -(a + b) \Rightarrow 2a + b = 0$ (Equation $1$)$b = ab$ (Equation $2$)From Equation $2$,$b(a - 1) = 0$,so $b = 0$ or $a = 1$.Case $1$: If $b = 0$,then from Equation $1$,$2a + 0 = 0 \Rightarrow a = 0$. Thus,$P(x) = x^2$. Then $P(2) = 2^2 = 4$.Case $2$: If $a = 1$,then from Equation $1$,$2(1) + b = 0 \Rightarrow b = -2$. Thus,$P(x) = x^2 + x - 2$. Then $P(2) = 2^2 + 2 - 2 = 4$.In both cases,$P(2) = 4$.

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

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