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

Please solve the following problem and show your work:

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

3 Answers

$$(ax + 2)(bx + 7) = 15x^2 + cx + 14$$ $$abx^2 + 7ax + 2bx + 14 = 15x^2 + cx + 14$$ Equating coefficients yields: $$ab = 15$$ $$7a + 2b = c$$ But $a + b = 8$. Therefore $a$ and $b$ are roots of the quadratic equation: $$y^2 - 8y + 15 = 0$$ $$(y - 5)(y - 3) = 0$$ Therefore, either $(a, b) = (3, 5)$ or $(a, b) = (5, 3)$.

Hence, either $$c = 7(3) + 2(5) = 31$$ or $$c = 7(5) + 2(3) = 41.$$

First we can expand the left hand side:

\begin{align}(ax+2)(bx+7)&=15x^2+cx+14\\ abx^2+2bx+7ax+14&=15x^2+cx+14\\ abx^2+(7a+2b)x+14&=15x^2+cx+14\end{align}

We can equate coefficients to see that

\begin{align}ab&=15\tag{$x^2$}\\ 7a+2b&=c\tag{$x$}\end{align}

We can say that \begin{align}a+b&=8\\a&=8-b\end{align}

And thus \begin{align}ab&=15\\ (8-b)b&=15\\ 8b-b^2&=15\\ b^2-8b+15&=0\\ (b-5)(b-3)&=0\end{align}

So we can say $b=3$ or $b=5$

When $b=3$, then $a=5$ and we can say that

\begin{align}c&=7a+2b\\ &=7\times 5+2\times 3\\ &= 35+6\\ &= 41\end{align}

When $b=5$, then $a=3$ and we can say that

\begin{align}c&=7a+2b\\ &=7\times 3+2\times 5\\ &= 21+10\\ &= 31\end{align}

So, the two values of $c$ are $41$ and $31$

It’s a simple problem. Just focus on the problem. You will find it easy.