Answer :
To solve for the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we need to match the coefficients of the polynomial on the left side of the equation with those on the right side. The polynomials given are:
[tex]\[
(x^3 + b x^2 - 7 x + 9)(x^2 + a x + 5) = x^5 + 13 x^4 + 38 x^3 - 22 x^2 + 37 x + 45
\][/tex]
We will expand the left-hand side and equate the coefficients with the corresponding terms on the right-hand side. Here are the steps:
1. Coefficient of [tex]\( x^5 \)[/tex]: This comes from multiplying the leading terms, [tex]\( x^3 \)[/tex] and [tex]\( x^2 \)[/tex], giving a coefficient of 1, which matches the right side. Therefore, there are no new conditions here.
2. Coefficient of [tex]\( x^4 \)[/tex]: This comes from:
- [tex]\( x^3 \cdot a x = a x^4 \)[/tex]
- [tex]\( b x^2 \cdot x^2 = b x^4 \)[/tex]
Adding these, the coefficient of [tex]\( x^4 \)[/tex] will be [tex]\( a + b \)[/tex].
Equate it to the coefficient on the right-hand side:
[tex]\[
a + b = 13
\][/tex]
3. Coefficient of [tex]\( x^3 \)[/tex]: This comes from:
- [tex]\( x^3 \cdot 5 = 5 x^3 \)[/tex]
- [tex]\( b x^2 \cdot a x = ba x^3 \)[/tex]
- [tex]\( -7 x \cdot x^2 = -7 x^3 \)[/tex]
Adding these, the coefficient of [tex]\( x^3 \)[/tex] is [tex]\( 5 + ba - 7 \)[/tex].
Equate it to the coefficient on the right-hand side:
[tex]\[
5 + ba - 7 = 38 \quad \Rightarrow \quad ba - 2 = 38 \quad \Rightarrow \quad ba = 40
\][/tex]
4. Coefficient of [tex]\( x^2 \)[/tex]: This comes from:
- [tex]\( b x^2 \cdot 5 = 5b x^2 \)[/tex]
- [tex]\( -7 x \cdot a x = -7a x^2 \)[/tex]
- [tex]\( 9 \cdot x^2 = 9 x^2 \)[/tex]
Adding these, the coefficient of [tex]\( x^2 \)[/tex] is [tex]\( 5b - 7a + 9 \)[/tex].
Equate it to the coefficient on the right-hand side:
[tex]\[
5b - 7a + 9 = -22 \quad \Rightarrow \quad 5b - 7a = -31
\][/tex]
Now, we have the system of equations:
1. [tex]\( a + b = 13 \)[/tex]
2. [tex]\( ba = 40 \)[/tex]
3. [tex]\( 5b - 7a = -31 \)[/tex]
Let's solve these equations:
- From equation 1, express [tex]\( a \)[/tex] in terms of [tex]\( b \)[/tex]:
[tex]\[
a = 13 - b
\][/tex]
- Substitute [tex]\( a = 13 - b \)[/tex] into the equation 2:
[tex]\[
b(13 - b) = 40 \quad \Rightarrow \quad 13b - b^2 = 40 \quad \Rightarrow \quad b^2 - 13b + 40 = 0
\][/tex]
This is a quadratic equation in [tex]\( b \)[/tex]. Solve for [tex]\( b \)[/tex]:
[tex]\[
b = \frac{13 \pm \sqrt{(13)^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1} = \frac{13 \pm \sqrt{169 - 160}}{2} = \frac{13 \pm 3}{2}
\][/tex]
This gives us two solutions for [tex]\( b \)[/tex]:
[tex]\[
b = 8 \quad \text{or} \quad b = 5
\][/tex]
- For [tex]\( b = 8 \)[/tex]:
[tex]\[
a = 13 - 8 = 5
\][/tex]
Check with equation 3, verify if [tex]\( 5b - 7a = -31 \)[/tex]:
[tex]\[
5(8) - 7(5) = 40 - 35 = 5 \quad \text{(which is not -31, so this pair is wrong)}
\][/tex]
- For [tex]\( b = 5 \)[/tex]:
[tex]\[
a = 13 - 5 = 8
\][/tex]
Check with equation 3:
[tex]\[
5(5) - 7(8) = 25 - 56 = -31 \quad \text{(which matches, so this pair is correct)}
\][/tex]
Thus, the solution to make the equation true is:
[tex]\[
a = 8, \quad b = 5
\][/tex]
[tex]\[
(x^3 + b x^2 - 7 x + 9)(x^2 + a x + 5) = x^5 + 13 x^4 + 38 x^3 - 22 x^2 + 37 x + 45
\][/tex]
We will expand the left-hand side and equate the coefficients with the corresponding terms on the right-hand side. Here are the steps:
1. Coefficient of [tex]\( x^5 \)[/tex]: This comes from multiplying the leading terms, [tex]\( x^3 \)[/tex] and [tex]\( x^2 \)[/tex], giving a coefficient of 1, which matches the right side. Therefore, there are no new conditions here.
2. Coefficient of [tex]\( x^4 \)[/tex]: This comes from:
- [tex]\( x^3 \cdot a x = a x^4 \)[/tex]
- [tex]\( b x^2 \cdot x^2 = b x^4 \)[/tex]
Adding these, the coefficient of [tex]\( x^4 \)[/tex] will be [tex]\( a + b \)[/tex].
Equate it to the coefficient on the right-hand side:
[tex]\[
a + b = 13
\][/tex]
3. Coefficient of [tex]\( x^3 \)[/tex]: This comes from:
- [tex]\( x^3 \cdot 5 = 5 x^3 \)[/tex]
- [tex]\( b x^2 \cdot a x = ba x^3 \)[/tex]
- [tex]\( -7 x \cdot x^2 = -7 x^3 \)[/tex]
Adding these, the coefficient of [tex]\( x^3 \)[/tex] is [tex]\( 5 + ba - 7 \)[/tex].
Equate it to the coefficient on the right-hand side:
[tex]\[
5 + ba - 7 = 38 \quad \Rightarrow \quad ba - 2 = 38 \quad \Rightarrow \quad ba = 40
\][/tex]
4. Coefficient of [tex]\( x^2 \)[/tex]: This comes from:
- [tex]\( b x^2 \cdot 5 = 5b x^2 \)[/tex]
- [tex]\( -7 x \cdot a x = -7a x^2 \)[/tex]
- [tex]\( 9 \cdot x^2 = 9 x^2 \)[/tex]
Adding these, the coefficient of [tex]\( x^2 \)[/tex] is [tex]\( 5b - 7a + 9 \)[/tex].
Equate it to the coefficient on the right-hand side:
[tex]\[
5b - 7a + 9 = -22 \quad \Rightarrow \quad 5b - 7a = -31
\][/tex]
Now, we have the system of equations:
1. [tex]\( a + b = 13 \)[/tex]
2. [tex]\( ba = 40 \)[/tex]
3. [tex]\( 5b - 7a = -31 \)[/tex]
Let's solve these equations:
- From equation 1, express [tex]\( a \)[/tex] in terms of [tex]\( b \)[/tex]:
[tex]\[
a = 13 - b
\][/tex]
- Substitute [tex]\( a = 13 - b \)[/tex] into the equation 2:
[tex]\[
b(13 - b) = 40 \quad \Rightarrow \quad 13b - b^2 = 40 \quad \Rightarrow \quad b^2 - 13b + 40 = 0
\][/tex]
This is a quadratic equation in [tex]\( b \)[/tex]. Solve for [tex]\( b \)[/tex]:
[tex]\[
b = \frac{13 \pm \sqrt{(13)^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1} = \frac{13 \pm \sqrt{169 - 160}}{2} = \frac{13 \pm 3}{2}
\][/tex]
This gives us two solutions for [tex]\( b \)[/tex]:
[tex]\[
b = 8 \quad \text{or} \quad b = 5
\][/tex]
- For [tex]\( b = 8 \)[/tex]:
[tex]\[
a = 13 - 8 = 5
\][/tex]
Check with equation 3, verify if [tex]\( 5b - 7a = -31 \)[/tex]:
[tex]\[
5(8) - 7(5) = 40 - 35 = 5 \quad \text{(which is not -31, so this pair is wrong)}
\][/tex]
- For [tex]\( b = 5 \)[/tex]:
[tex]\[
a = 13 - 5 = 8
\][/tex]
Check with equation 3:
[tex]\[
5(5) - 7(8) = 25 - 56 = -31 \quad \text{(which matches, so this pair is correct)}
\][/tex]
Thus, the solution to make the equation true is:
[tex]\[
a = 8, \quad b = 5
\][/tex]
