If $(ax+2)(bx+7)=15x^2+cx+14$ for all values of $x$, and $a+b=8$, what are two possible values for $c$? [closed]
Solution 1:
We put it as
$$(ax + 2) (bx +7) = 15x^2 + cx +14$$
$$abx^2 +(7a+2b)x + 14 = 15x^2 + cx +14$$
By comparison we take $ab=15$ and $a+b=8$ and solving them we get two cases:
1- $a=3$ and $b=5$, and $c= 7a+2b = 31$.
2- $a=5$ and $b=3$, and $c= 7a+2b = 40$