Question

If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c

Answer 1

Given:

$(ax+2)(bx+7)=15x^2+cx+14$

And

$a+b=8$

Required:

To find the two possible values of c.

Explanation:

Consider

$\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}$

So

$\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}$

And also given

$a+b=8---(2)$

Now from (1) and (2), we get

$\begin{gathered} a+\frac{15}{a}=8 \\ \\ a^2+15=8a \\ \\ a^2-8a+15=0 \end{gathered}$$a=3,5$

Now put a in (1) we get

$\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}$

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

$\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}$

When a=5 and b=3, we have

$\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}$

Final Answer:

The option D is correct.

31 and 41