Question

Find the solutions to the equation 4x^4 −8x^3−41x^2+18x+72=0 if x+2 and 2x−3 are factors of the polynomial 4x4 −8x^3−41x^2+18x+72.

Answer 1

To solve the equation \( 4x^4 - 8x^3 - 41x^2 + 18x + 72 = 0 \) given that \( x + 2 \) and \( 2x - 3 \) are factors, we'll first find the corresponding roots for these factors.

### Step 1: Find the roots of the factors
The roots corresponding to the factors are:
1. For \( x + 2 = 0 \):
\[
x = -2
\]
2. For \( 2x - 3 = 0 \):
\[
2x = 3 \implies x = \frac{3}{2}
\]

Thus, we have two roots: \( x = -2 \) and \( x = \frac{3}{2} \).

### Step 2: Polynomial Division
Since \( x + 2 \) and \( 2x - 3 \) are factors, we can perform polynomial long division to divide the original polynomial by the factors.

#### Dividing by \( x + 2 \)
1. Divide \( 4x^4 - 8x^3 - 41x^2 + 18x + 72 \) by \( x + 2 \).

The first term of the quotient is \( 4x^3 \) since \( 4x^3(x + 2) = 4x^4 + 8x^3 \).
Subtract this from the original polynomial:
\[
(4x^4 - 8x^3) - (4x^4 + 8x^3) = -16x^3
\]
Now, bring down the next term:
\[
-16x^3 - 41x^2 = -16x^3 - 41x^2
\]
The next term is \( -16x^2 \) since \( -16x^2(x + 2) = -16x^3 - 32x^2 \). Subtract again:
\[
(-16x^3 - 41x^2) - (-16x^3 - 32x^2) = -9x^2
\]
Bring down the next term:
\[
-9x^2 + 18x = -9x^2 + 18x
\]
The next term is \( -9x \) since \( -9x(x + 2) = -9x^2 - 18x \). Subtract:
\[
(-9x^2 + 18x) - (-9x^2 - 18x) = 36x
\]
Finally, bring down the last term:
\[
36x + 72
\]
The next term is \( 36 \) since \( 36(x + 2) = 36x + 72 \). Subtract:
\[
(36x + 72) - (36x + 72) = 0
\]

Thus, the result of this division is:
\[
4x^3 - 16x^2 - 9x + 36
\]

#### Dividing by \( 2x - 3 \)
Now we divide \( 4x^3 - 16x^2 - 9x + 36 \) by \( 2x - 3 \).

1. The first term of the quotient is \( 2x^2 \).
\( 2x^2(2x - 3) = 4x^3 - 6x^2 \).
Subtract:
\[
(-16x^2) - (-6x^2) = -10x^2
\]
Bring down the next term:
\[
-10x^2 - 9x = -10x^2 - 9x
\]
The next term is \( -5x \) because \( -5x(2x - 3) = -10x^2 + 15x \), Subtract:
\[
(-10x^2 - 9x) - (-10x^2 + 15x) = -24x
\]
Lastly, bring down the \( +36 \):
\[
-24x + 36
\]
The next term is \( -12 \) since \( -12(2x - 3) = -24x + 36 \). Subtract:
\[
(-24x + 36) - (-24x + 36) = 0
\]

So we are left with:
\[
2x^2 - 5x - 12
\]

### Step 3: Factor the quadratic
Now we will factor \( 2x^2 - 5x - 12 \).

1. To factor, we need two numbers that multiply to \( 2 \times (-12) = -24 \) and add to \( -5 \). These numbers are \( -8 \) and \( 3 \).
2. Rewrite:
\[
2x^2 - 8x + 3x - 12
\]
3. Factor by grouping:
\[
2x(x - 4) + 3(x - 4) = (2x + 3)(x - 4)
\]

### Step 4: Final roots
Now the complete factorization of the polynomial is:
\[
(x + 2)(2x - 3)(2x + 3)(x - 4) = 0
\]

Setting each factor to zero gives the roots:
1. \( x + 2 = 0 \rightarrow x = -2 \)
2. \( 2x - 3 = 0 \rightarrow x = \frac{3}{2} \)
3. \( 2x + 3 = 0 \rightarrow x = -\frac{3}{2} \)
4. \( x - 4 = 0 \rightarrow x = 4 \)

### Summary of solutions:
The solutions to the equation \( 4x^4 - 8x^3 - 41x^2 + 18x + 72 = 0 \) are:
\[
x = -2, \quad x = \frac{3}{2}, \quad x = -\frac{3}{2}, \quad x = 4
\]