To solve this problem, we need to expand the left-hand side of the given equation [tex]\((a x + 2)(b x + 7)\)[/tex] and compare it to the right-hand side [tex]\(15 x^2 + c x + 14\)[/tex].
Let's expand the left side step-by-step:
[tex]\[
(a x + 2)(b x + 7)
\][/tex]
Using the distributive property (FOIL method):
[tex]\[
(a x + 2)(b x + 7) = a b x^2 + 7 a x + 2 b x + 14
\][/tex]
Combining the [tex]\(x\)[/tex] terms:
[tex]\[
= a b x^2 + (7 a + 2 b) x + 14
\][/tex]
Next, we'll match the coefficients from the left side to the right side of the equation [tex]\(15 x^2 + c x + 14\)[/tex]:
1. Coefficient of [tex]\(x^2\)[/tex]:
[tex]\[
a b = 15
\][/tex]
2. Coefficient of [tex]\(x\)[/tex]:
[tex]\[
7 a + 2 b = c
\][/tex]
3. Constant term:
[tex]\[
14 = 14 \quad (\text{This is automatically satisfied.})
\][/tex]
We are given the additional relation [tex]\(a + b = 8\)[/tex]. Now we need to solve for [tex]\(a\)[/tex] and [tex]\(b\)[/tex] using the equations [tex]\(a b = 15\)[/tex] and [tex]\(a + b = 8\)[/tex].
We have two simultaneous equations:
1. [tex]\(a b = 15\)[/tex]
2. [tex]\(a + b = 8\)[/tex]
We solve these equations to find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
[tex]\[
a + b = 8 \implies b = 8 - a
\][/tex]
Substitute [tex]\(b\)[/tex] into the first equation:
[tex]\[
a (8 - a) = 15 \implies 8a - a^2 = 15 \implies a^2 - 8a + 15 = 0
\][/tex]
This is a quadratic equation. Solve it using the quadratic formula [tex]\(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -8\)[/tex], and [tex]\(c = 15\)[/tex]:
[tex]\[
a = \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm \sqrt{4}}{2} = \frac{8 \pm 2}{2}
\][/tex]
This gives us two values for [tex]\(a\)[/tex]:
[tex]\[
a = \frac{8 + 2}{2} = 5 \quad \text{and} \quad a = \frac{8 - 2}{2} = 3
\][/tex]
Corresponding [tex]\(b\)[/tex] values are:
[tex]\[
\text{If } a = 5, \text{ then } b = 8 - 5 = 3
\][/tex]
[tex]\[
\text{If } a = 3, \text{ then } b = 8 - 3 = 5
\][/tex]
So the pairs [tex]\((a, b)\)[/tex] that satisfy the conditions are [tex]\((5, 3)\)[/tex] and [tex]\((3, 5)\)[/tex].
Next, we use these pairs to find the two possible values of [tex]\(c\)[/tex]:
1. For [tex]\(a = 5\)[/tex] and [tex]\(b = 3\)[/tex]:
[tex]\[
7a + 2b = 7(5) + 2(3) = 35 + 6 = 41
\][/tex]
2. For [tex]\(a = 3\)[/tex] and [tex]\(b = 5\)[/tex]:
[tex]\[
7a + 2b = 7(3) + 2(5) = 21 + 10 = 31
\][/tex]
Thus, the two possible values for [tex]\(c\)[/tex] are:
[tex]\[
\boxed{31 \text{ and } 41}
\][/tex]
