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Answer :
Sure, let's solve the division of the polynomials [tex]\(\frac{6x^3 + 19x^2 - 20}{2x + 1}\)[/tex].
Step-by-Step Solution:
1. Set up the division:
We are dividing [tex]\(6x^3 + 19x^2 - 20\)[/tex] by [tex]\(2x + 1\)[/tex].
2. Divide the first term of the numerator by the first term of the denominator:
[tex]\[
\frac{6x^3}{2x} = 3x^2
\][/tex]
3. Multiply the entire denominator by this result:
[tex]\[
3x^2 \cdot (2x + 1) = 6x^3 + 3x^2
\][/tex]
4. Subtract this from the original polynomial:
[tex]\[
(6x^3 + 19x^2 - 20) - (6x^3 + 3x^2) = 16x^2 - 20
\][/tex]
5. Repeat the process with the new polynomial (16x^2 - 20):
[tex]\[
\frac{16x^2}{2x} = 8x
\][/tex]
Multiply the denominator:
[tex]\[
8x \cdot (2x + 1) = 16x^2 + 8x
\][/tex]
Subtract:
[tex]\[
(16x^2 - 20) - (16x^2 + 8x) = -8x - 20
\][/tex]
6. Repeat again with (-8x - 20):
[tex]\[
\frac{-8x}{2x} = -4
\][/tex]
Multiply the denominator:
[tex]\[
-4 \cdot (2x + 1) = -8x - 4
\][/tex]
Subtract:
[tex]\[
(-8x - 20) - (-8x - 4) = -16
\][/tex]
7. Combine the terms to get the quotient and remainder:
The quotient is [tex]\(3x^2 + 8x - 4\)[/tex].
The remainder is [tex]\(-16\)[/tex].
Thus, the result of the division [tex]\(\frac{6x^3 + 19x^2 - 20}{2x + 1}\)[/tex] is:
[tex]\[
3x^2 + 8x - 4 \quad \text{with a remainder of} \quad -16
\][/tex]
Step-by-Step Solution:
1. Set up the division:
We are dividing [tex]\(6x^3 + 19x^2 - 20\)[/tex] by [tex]\(2x + 1\)[/tex].
2. Divide the first term of the numerator by the first term of the denominator:
[tex]\[
\frac{6x^3}{2x} = 3x^2
\][/tex]
3. Multiply the entire denominator by this result:
[tex]\[
3x^2 \cdot (2x + 1) = 6x^3 + 3x^2
\][/tex]
4. Subtract this from the original polynomial:
[tex]\[
(6x^3 + 19x^2 - 20) - (6x^3 + 3x^2) = 16x^2 - 20
\][/tex]
5. Repeat the process with the new polynomial (16x^2 - 20):
[tex]\[
\frac{16x^2}{2x} = 8x
\][/tex]
Multiply the denominator:
[tex]\[
8x \cdot (2x + 1) = 16x^2 + 8x
\][/tex]
Subtract:
[tex]\[
(16x^2 - 20) - (16x^2 + 8x) = -8x - 20
\][/tex]
6. Repeat again with (-8x - 20):
[tex]\[
\frac{-8x}{2x} = -4
\][/tex]
Multiply the denominator:
[tex]\[
-4 \cdot (2x + 1) = -8x - 4
\][/tex]
Subtract:
[tex]\[
(-8x - 20) - (-8x - 4) = -16
\][/tex]
7. Combine the terms to get the quotient and remainder:
The quotient is [tex]\(3x^2 + 8x - 4\)[/tex].
The remainder is [tex]\(-16\)[/tex].
Thus, the result of the division [tex]\(\frac{6x^3 + 19x^2 - 20}{2x + 1}\)[/tex] is:
[tex]\[
3x^2 + 8x - 4 \quad \text{with a remainder of} \quad -16
\][/tex]
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