Question

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

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

Answer 1

`(ax+2)(bx+7)=abx^2+(7a+2b)x+14=15x^2+cx+14`

For the two quadratics to match, the coefficients of corresponding terms must be equal, so that

`\begin{cases}ab=15\\7a+2b=c\end{cases}`

From the given options, we are only allowing
`c` to be an integer, which means
`a,b` must also be integers. Then we have eight possible choices for
`a,b`:

1, 15

3, 5

5, 3

15, 1

plus the same combinations with negative signs on both numbers. But since the choices for
`c` are also positive, we can ignore the negative versions.

If
`(a,b)=(1,15)`, then
`7a+2b=37`.

If
`(a,b)=(3,5)`, then
`7a+2b=31`.

If
`(a,b)=(5,3)`, then
`7a+2b=41`.

If
`(a,b)=(15,1)`, then
`7a+2b=107`.

So the answer is C.

Answer 2

The two possible values of
`C` is option
`C`
`(31` and
`41)`.

Given,

`(ax+2)(bx+7)=15x^2+cx+14\\a+b=8`

Simplify the equation of LHS as,

`(ax+2)(bx+7)=abx^2+2bx+7ax+14\\=abx^2+x(7a+2b)+14`

Compare LHS and RHS as,

`abx^2+x(7a+2b)+14=15x^2+cx+14\\abx^2+(7a+2b)x=15x^2+cx`

The two equations are:

`ab=15\\7a+2b=c`

From the given,

`a+b=8`

Subtract
`b` on both sides in the equation as,

`a+b-b=8-b\\a=8-b`

Substitute
`a=8-b` in the equation
`ab=15` as,

`(8-b)b=15\\8b-b^2=15\\-b^2+8b=15\\`

Solve the equation by setting equal to zero as,

`-b^2+8b-15=0\\b^2-8b+15=0`

Factorize the equation as follows:

`b^2--8b+15=0\\b^2-5b-3b+15=0\\b(b-5)-3(b-5)=0\\(b-3)(b-5)=0`

Solve the factors as,

`b-3=0\\b=3`

And

`b-5=0\\b=5`

Substitute
`b=3` in
`a=8-b` as,

`a=8-3\\=5`

Thus,
`a=5,b=3`.

Substitute
`b=5` in
`a=8-b` as,

`a=8-5\\=3`

Thus,
`a=3,b=5`.

Substitute
`a=3,b=5` in the equation
`7a+2b=c` as,

`7(3)+2(5)=c\\21+10=c\\c=31`

Substitute
`a=5,b=3` in the equation
`7a+2b=c` as,

`7(5)+2(3)=c\\35+6=c\\c=41`

Thus, the two possible values of
`C` is
`31` and
`41`.