PSAT #1, Section 3, #17 – PWN Test Prep

PSAT #1, Section 3, #17

If $x-2$ is a factor of $x^2-bx+b$ , then you know that for some other factor $x-a$ ,

$\begin{align*}(x-2)(x-a)&=x^2-bx+b\\x^2-2x-ax+2a&=x^2-bx+b\end{align*}$

The equivalent polynomials rule tells you that in this case, $2a=b$ and $-2x-ax=-bx$ . Simplify that second one a bit and you get $2+a=b$ . You might just see the answer at this point, but you can also solve the system by substituting for $b$ :

$\begin{align*}2+a&=2a\\2&=a\end{align*}$

If $a=2$ , then $b=4$ .

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