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If $(a x+2)(b x+7)=15 x^{2}+c x+14$ for all values of $x$, and $a+b=8$, what are the two possible values for $c$ ?

SAT · Math · previous-year question

  1. A.3 and 5
  2. B.6 and 35
  3. C.10 and 21
  4. D.31 and 41correct

Answer

D. 31 and 41

Explanation

Choice D is correct. One can find the possible values of $a$ and $b$ in $(a x+2)(b x+7)$ by using the given equation $a+b=8$ and finding another equation that relates the variables $a$ and $b$. Since $(a x+2)(b x+7)=15 x^{2}+c x+14$, one can expand the left side of the equation to obtain $a b x^{2}+7 a x+2 b x+14=15 x^{2}+c x+14$. Since $a b$ is the coefficient of $x^{2}$ on the left side of the equation and 15 is the coefficient of $x^{2}$ on the right side of the equation, it must be true that $a b=15$. Since $a+b=8$, it follows that $b=8-a$. Thus, $a b=15$ can be rewritten as $a 8-a)=15$, w

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