Given f(x) = ax^2 + bx + c,g(x) = a_1x^2 + b_ | vedclass

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(A) Since $p(x) = f(x) - g(x)$,we have $p(x) = (a - a_1)x^2 + (b - b_1)x + (c - c_1)$.Let $A = a - a_1$,$B = b - b_1$,and $C = c - c_1$. Then $p(x) =

Vedclass · Accepted Answer

$18$

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(A) Since $p(x) = f(x) - g(x)$,we have $p(x) = (a - a_1)x^2 + (b - b_1)x + (c - c_1)$.Let $A = a - a_1$,$B = b - b_1$,and $C = c - c_1$. Then $p(x) = Ax^2 + Bx + C$.Given that $p(x) = 0$ only for $x = -1$,it implies that $x = -1$ is a repeated root of the quadratic equation $p(x) = 0$.Therefore,$p(x) = A(x + 1)^2$.We are given $p(-2) = 2$.Substituting $x = -2$ into the expression: $p(-2) = A(-2 + 1)^2 = A(-1)^2 = A$.Thus,$A = 2$.Now,we need to find $p(2)$:$p(2) = A(2 + 1)^2 = 2(3)^2 = 2(9) = 18$.

Given $f(x) = ax^2 + bx + c$,$g(x) = a_1x^2 + b_1x + c_1$,and $p(x) = f(x) - g(x)$. If $p(x) = 0$ only for $x = -1$ and $p(-2) = 2$,what is the value of $p(2)$? Assume $a \neq a_1 \neq 0$.

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