Answer :
To solve this problem, we will first expand the left-hand side of the given equation [tex]\((a x+2)(b x+3)\)[/tex] and then match the coefficients with the right-hand side [tex]\(15 x^2 + c x + 6\)[/tex].
Let's start by expanding [tex]\((a x + 2)(b x + 3)\)[/tex]:
[tex]\[ (a x + 2)(b x + 3) = (a x)(b x) + (a x)(3) + (2)(b x) + (2)(3) \][/tex]
[tex]\[ = ab x^2 + 3a x + 2b x + 6 \][/tex]
Combine the like terms [tex]\(3a x + 2b x\)[/tex]:
[tex]\[ = ab x^2 + (3a + 2b) x + 6 \][/tex]
Now, we want this to be equal to [tex]\(15 x^2 + c x + 6\)[/tex] for all values of [tex]\(x\)[/tex]. Therefore, we can equate the coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant term from both sides of the equation.
By comparing coefficients we get:
[tex]\[ ab = 15 \][/tex]
[tex]\[ 3a + 2b = c \][/tex]
[tex]\[ 6 = 6 \][/tex]
The last equation (constant term) is always true and doesn't provide us any new information. We are left with:
1. [tex]\(ab = 15\)[/tex]
2. [tex]\(3a + 2b = c\)[/tex]
3. [tex]\(a + b = 8\)[/tex]
We need to find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that satisfy both [tex]\(ab = 15\)[/tex] and [tex]\(a + b = 8\)[/tex].
Consider the quadratic equation whose roots are [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ t^2 - (a + b)t + ab = 0 \][/tex]
Substitute [tex]\(a + b = 8\)[/tex] and [tex]\(ab = 15\)[/tex]:
[tex]\[ t^2 - 8t + 15 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\(t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)[/tex]:
[tex]\[ t = \frac{8 \pm \sqrt{64 - 60}}{2} \][/tex]
[tex]\[ t = \frac{8 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ t = \frac{8 \pm 2}{2} \][/tex]
[tex]\[ t = \frac{8 + 2}{2} = 5 \quad \text{or} \quad t = \frac{8 - 2}{2} = 3 \][/tex]
So, the possible values for [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are 5 and 3, respectively.
Now, substitute these values back into [tex]\(3a + 2b = c\)[/tex]:
1. If [tex]\(a = 5\)[/tex] and [tex]\(b = 3\)[/tex]:
[tex]\[ c = 3(5) + 2(3) = 15 + 6 = 21 \][/tex]
2. If [tex]\(a = 3\)[/tex] and [tex]\(b = 5\)[/tex]:
[tex]\[ c = 3(3) + 2(5) = 9 + 10 = 19 \][/tex]
Therefore, the two possible values for [tex]\(c\)[/tex] are 19 and 21.
So, the correct answer is:
[tex]\[ \boxed{(C) 19, 21} \][/tex]
Let's start by expanding [tex]\((a x + 2)(b x + 3)\)[/tex]:
[tex]\[ (a x + 2)(b x + 3) = (a x)(b x) + (a x)(3) + (2)(b x) + (2)(3) \][/tex]
[tex]\[ = ab x^2 + 3a x + 2b x + 6 \][/tex]
Combine the like terms [tex]\(3a x + 2b x\)[/tex]:
[tex]\[ = ab x^2 + (3a + 2b) x + 6 \][/tex]
Now, we want this to be equal to [tex]\(15 x^2 + c x + 6\)[/tex] for all values of [tex]\(x\)[/tex]. Therefore, we can equate the coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant term from both sides of the equation.
By comparing coefficients we get:
[tex]\[ ab = 15 \][/tex]
[tex]\[ 3a + 2b = c \][/tex]
[tex]\[ 6 = 6 \][/tex]
The last equation (constant term) is always true and doesn't provide us any new information. We are left with:
1. [tex]\(ab = 15\)[/tex]
2. [tex]\(3a + 2b = c\)[/tex]
3. [tex]\(a + b = 8\)[/tex]
We need to find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that satisfy both [tex]\(ab = 15\)[/tex] and [tex]\(a + b = 8\)[/tex].
Consider the quadratic equation whose roots are [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ t^2 - (a + b)t + ab = 0 \][/tex]
Substitute [tex]\(a + b = 8\)[/tex] and [tex]\(ab = 15\)[/tex]:
[tex]\[ t^2 - 8t + 15 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\(t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)[/tex]:
[tex]\[ t = \frac{8 \pm \sqrt{64 - 60}}{2} \][/tex]
[tex]\[ t = \frac{8 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ t = \frac{8 \pm 2}{2} \][/tex]
[tex]\[ t = \frac{8 + 2}{2} = 5 \quad \text{or} \quad t = \frac{8 - 2}{2} = 3 \][/tex]
So, the possible values for [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are 5 and 3, respectively.
Now, substitute these values back into [tex]\(3a + 2b = c\)[/tex]:
1. If [tex]\(a = 5\)[/tex] and [tex]\(b = 3\)[/tex]:
[tex]\[ c = 3(5) + 2(3) = 15 + 6 = 21 \][/tex]
2. If [tex]\(a = 3\)[/tex] and [tex]\(b = 5\)[/tex]:
[tex]\[ c = 3(3) + 2(5) = 9 + 10 = 19 \][/tex]
Therefore, the two possible values for [tex]\(c\)[/tex] are 19 and 21.
So, the correct answer is:
[tex]\[ \boxed{(C) 19, 21} \][/tex]
