Question

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

• A. $3$ and $5$
• B. $6$ and $35$
• C. $10$ and $21$
• D. $31$ and $41$

Answer 1

The correct option is D $31$ and $41$

We have $(ax+2)(bx+7)=ab{x}^2+7ax+2bx+14=ab{x}^2+(7a+2b)x+14$
Given, it is equal to $15x^2+cx+14$
$\Rightarrow ab=15$ --- (1)
$\Rightarrow 7a+2b=c$
Also given $a+b=8$ --- (2)
From equations (1) and (2), we can interpret that either of $a$ or $b$ are equal to $3$ or $5$
When $a=3$ and $b=5$, we have $c=7\left(3\right)+2\left(5\right)=21+10=31$
When $a=5$ and $b=3$, we have $c=7\left(5\right)+2\left(3\right)=35+6=41$