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Show by actual division that [tex]\(2x + 3\)[/tex] is a factor of [tex]\(f(x) = 4x^4 + 8x^3 + 5x^2 + x - 3\)[/tex].

Sagot :

To determine if [tex]\(2x + 3\)[/tex] is a factor of [tex]\(f(x) = 4x^4 + 8x^3 + 5x^2 + x - 3\)[/tex] by actual division, we will perform polynomial long division. The steps are as follows:

1. Set up the division: Place the dividend [tex]\(4x^4 + 8x^3 + 5x^2 + x - 3\)[/tex] inside the division bracket and the divisor [tex]\(2x + 3\)[/tex] outside.

2. First Term: Divide the leading term of the dividend [tex]\(4x^4\)[/tex] by the leading term of the divisor [tex]\(2x\)[/tex]:
[tex]\[ \frac{4x^4}{2x} = 2x^3 \][/tex]

3. Multiply and Subtract: Multiply [tex]\(2x^3\)[/tex] by [tex]\(2x + 3\)[/tex] and subtract from the original polynomial:
[tex]\[ (2x^3) \cdot (2x + 3) = 4x^4 + 6x^3 \][/tex]
[tex]\[ (4x^4 + 8x^3 + 5x^2 + x - 3) - (4x^4 + 6x^3) = 2x^3 + 5x^2 + x - 3 \][/tex]

4. Second Term: Repeat the process with the new leading term [tex]\(2x^3\)[/tex]:
[tex]\[ \frac{2x^3}{2x} = x^2 \][/tex]
Multiply [tex]\(x^2\)[/tex] by [tex]\(2x + 3\)[/tex] and subtract:
[tex]\[ (x^2) \cdot (2x + 3) = 2x^3 + 3x^2 \][/tex]
[tex]\[ (2x^3 + 5x^2 + x - 3) - (2x^3 + 3x^2) = 2x^2 + x - 3 \][/tex]

5. Third Term: Proceed with the next leading term [tex]\(2x^2\)[/tex]:
[tex]\[ \frac{2x^2}{2x} = x \][/tex]
Multiply [tex]\(x\)[/tex] by [tex]\(2x + 3\)[/tex] and subtract:
[tex]\[ (x) \cdot (2x + 3) = 2x^2 + 3x \][/tex]
[tex]\[ (2x^2 + x - 3) - (2x^2 + 3x) = -2x - 3 \][/tex]

6. Fourth Term: Continue with [tex]\(-2x\)[/tex]:
[tex]\[ \frac{-2x}{2x} = -1 \][/tex]
Multiply [tex]\(-1\)[/tex] by [tex]\(2x + 3\)[/tex] and subtract:
[tex]\[ (-1) \cdot (2x + 3) = -2x - 3 \][/tex]
[tex]\[ (-2x - 3) - (-2x - 3) = 0 \][/tex]

The remainder is zero.

The quotient is given by combining the terms obtained in each step:
[tex]\[ 2x^3 + x^2 + x - 1 \][/tex]
and the remainder is 0.

Thus, [tex]\(2x + 3\)[/tex] is indeed a factor of [tex]\(f(x) = 4x^4 + 8x^3 + 5x^2 + x - 3\)[/tex], as evidenced by the quotient [tex]\(2x^3 + x^2 + x - 1\)[/tex] and a remainder of 0.