Question

Find the solutions to the equation below if x + 2 and 2x - 3 are factors of the polynomial 4x^4 -8x^3 - 41x^2 + 18x + 72

Answer 1

To find the solutions to the polynomial \(4x^4 - 8x^3 - 41x^2 + 18x + 72\) given that \(x + 2\) and \(2x - 3\) are factors, we can start by using the Factor Theorem.

1. **Find the roots of the factors**:
- For the factor \(x + 2\):
\[ x + 2 = 0 \Rightarrow x = -2 \]
- For the factor \(2x - 3\):
\[ 2x - 3 = 0 \Rightarrow x = \frac{3}{2} \]

2. **Verify these roots are indeed roots of the polynomial**:
- Substitute \(x = -2\) into the polynomial:

\[
P(-2) = 4(-2)^4 - 8(-2)^3 - 41(-2)^2 + 18(-2) + 72
= 4(16) + 8(8) - 41(4) - 36 + 72
= 64 + 64 - 164 - 36 + 72
= 0
\]
- Therefore, \(x = -2\) is a root.

- Substitute \(x = \frac{3}{2}\) into the polynomial:

\[
P\left(\frac{3}{2}\right) = 4\left(\frac{3}{2}\right)^4 - 8\left(\frac{3}{2}\right)^3 - 41\left(\frac{3}{2}\right)^2 + 18\left(\frac{3}{2}\right) + 72
\]
\[
= 4\left(\frac{81}{16}\right) - 8\left(\frac{27}{8}\right) - 41\left(\frac{9}{4}\right) + 27 + 72
= \frac{324}{16} - \frac{216}{8} - \frac{369}{4} + 27 + 72
= \frac{324}{16} - \frac{432}{16} - \frac{1476}{16} + \frac{432}{16} + \frac{1152}{16}
= 0
\]
- Therefore, \(x = \frac{3}{2}\) is also a root.

3. **Use polynomial long division to factor the polynomial**:
Since we know \(x + 2\) and \(2x - 3\) are factors, we can divide the polynomial by the product of these two factors. First, we can multiply these factors together:

\[
(x + 2)(2x - 3) = 2x^2 + 4x - 3
\]

4. **Divide the polynomial \(P(x)\) by \(2x^2 + 4x - 3\)**.

Perform polynomial long division of:

\[
4x^4 - 8x^3 - 41x^2 + 18x + 72 \div (2x^2 + 4x - 3)
\]

1. **First term**: Divide the leading term:
\[
4x^4 \div 2x^2 = 2x^2
\]
Multiply \(2x^2\) with \(2x^2 + 4x - 3\):
\[
4x^4 + 8x^3 - 6x^2
\]
Subtract:
\[
(4x^4 - 8x^3 - 41x^2) - (4x^4 + 8x^3 - 6x^2) = -16x^3 - 35x^2
\]

2. **Next term**: Divide leading term:
\[
-16x^3 \div 2x^2 = -8x
\]
Multiply \(-8x\) with \(2x^2 + 4x - 3\):
\[
-16x^3 - 32x^2 + 24x
\]
Subtract:
\[
(-16x^3 - 35x^2 + 18x) - (-16x^3 - 32x^2 + 24x) = -3x^2 - 6x
\]

3. **Next term**: Divide leading term:
\[
-3x^2 \div 2x^2 = -\frac{3}{2}
\]
Multiply \(-\frac{3}{2}\) with \(2x^2 + 4x - 3\):
\[
-3x^2 - 6x + \frac{9}{2}
\]
Subtract:
\[
(-3x^2 - 6x + 72) - (-3x^2 - 6x + \frac{9}{2}) = 72 - \frac{9}{2} = \frac{144}{2} - \frac{9}{2} = \frac{135}{2}
\]

Thus, the polynomial can be factored as follows:

\[
P(x) = (2x^2 + 4x - 3)(2x^2 - 1) = (x + 2)(2x - 3)(2x^2 - 1)
\]

Now factor \(2x^2 - 1\):

\[
2x^2 - 1 = 0 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}
\]

5. **Final solutions**:
The solutions to the original polynomial are:
\[
x = -2, \quad x = \frac{3}{2}, \quad x = \frac{\sqrt{2}}{2}, \quad x = -\frac{\sqrt{2}}{2}
\]