Question

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

Answer 1

The two possible values for c are 31 and 41

Step-by-step explanation:

Given Expression:

$(a x+2)(b x+7)=15 x^{2}+c x+14$

a + b = 8

To expand,

Multiply a x with (b x + 7) =  $a b x^{2}+7 a x$

Multiply 2 with (b x + 7) = 2 b x +14

Now, combining the above, we get

         $a b x^{2}+7 a x+2 b x+14=15 x^{2}+c x+14$

         $a b x^{2}+x(7 a+2 b)+14=a b x^{2}$

When comparing both sides, we get

a b = 15, 7 a + 2 b = c

          $a=\frac{15}{b} \text { or } b=\frac{a}{15}$

Now, substitute above value in a + b = 8. So,

         $\frac{15}{b}+b=8$

         $15+b^{2}=8 b$

        $b^{2}-8 b+15=0$

Factorising above, we get the equation as

       $b^{2}-3 b-5 b+15=0$

       (b - 3) (b - 5) = 0

b = 3 and 5

If b = 3, then

       $a=\frac{15}{b}=\frac{15}{3}=5$

If b = 5, then

       $a=\frac{15}{5}=3$

If a = 3, b = 5

c =7 a + 2 b = 7 (3) + 2 (5) = 21 + 10 = 31

If a = 5, b = 3

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

Therefore, the values of ‘c’ are 31 and 41 .