Sure, let’s demonstrate step-by-step that [tex]\(2x + 3\)[/tex] is a factor of [tex]\(f(x) = 4x^4 + 8x^3 + 5x^2 + x - 3\)[/tex] by performing polynomial division.
1. Setup the division:
We need to divide [tex]\(4x^4 + 8x^3 + 5x^2 + x - 3\)[/tex] by [tex]\(2x + 3\)[/tex].
2. First division:
Divide the leading term [tex]\(4x^4\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[
\frac{4x^4}{2x} = 2x^3
\][/tex]
Multiply [tex]\(2x^3\)[/tex] by [tex]\(2x + 3\)[/tex]:
[tex]\[
2x^3 \cdot (2x + 3) = 4x^4 + 6x^3
\][/tex]
Subtract from the original polynomial:
[tex]\[
(4x^4 + 8x^3 + 5x^2 + x - 3) - (4x^4 + 6x^3) = 2x^3 + 5x^2 + x - 3
\][/tex]
3. Second division:
Divide the new leading term [tex]\(2x^3\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[
\frac{2x^3}{2x} = x^2
\][/tex]
Multiply [tex]\(x^2\)[/tex] by [tex]\(2x + 3\)[/tex]:
[tex]\[
x^2 \cdot (2x + 3) = 2x^3 + 3x^2
\][/tex]
Subtract:
[tex]\[
(2x^3 + 5x^2 + x - 3) - (2x^3 + 3x^2) = 2x^2 + x - 3
\][/tex]
4. Third division:
Divide the new leading term [tex]\(2x^2\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[
\frac{2x^2}{2x} = x
\][/tex]
Multiply [tex]\(x\)[/tex] by [tex]\(2x + 3\)[/tex]:
[tex]\[
x \cdot (2x + 3) = 2x^2 + 3x
\][/tex]
Subtract:
[tex]\[
(2x^2 + x - 3) - (2x^2 + 3x) = -2x - 3
\][/tex]
5. Fourth division:
Divide the new leading term [tex]\(-2x\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[
\frac{-2x}{2x} = -1
\][/tex]
Multiply [tex]\(-1\)[/tex] by [tex]\(2x + 3\)[/tex]:
[tex]\[
-1 \cdot (2x + 3) = -2x - 3
\][/tex]
Subtract:
[tex]\[
(-2x - 3) - (-2x - 3) = 0
\][/tex]
6. Conclusion:
The remainder after the polynomial division is 0, which confirms that [tex]\(2x + 3\)[/tex] is indeed a factor of [tex]\(4x^4 + 8x^3 + 5x^2 + x - 3\)[/tex]. Hence, the quotient of the division is [tex]\(2x^3 + x^2 + x - 1\)[/tex].
