To apply polynomial long division using Horner's method for the given polynomials, let us divide the polynomial [tex]\(15x^6 - 9x^5 + 56x^4 + 70x^3 + 73x - 23\)[/tex] by [tex]\(3x^2 + 7x + 11\)[/tex].
### Step-by-Step Solution:
1. Identify the leading terms and divide:
- The leading term of the numerator is [tex]\(15x^6\)[/tex].
- The leading term of the denominator is [tex]\(3x^2\)[/tex].
- Divide these leading terms: [tex]\(\frac{15x^6}{3x^2} = 5x^4\)[/tex].
2. Multiply the entire divisor by this result:
- [tex]\(5x^4 \cdot (3x^2 + 7x + 11) = 15x^6 + 35x^5 + 55x^4\)[/tex].
3. Subtract this result from the original numerator:
[tex]\[
(15x^6 - 9x^5 + 56x^4 + 70x^3 + 73x - 23) - (15x^6 + 35x^5 + 55x^4) = -44x^5 + x^4 + 70x^3 + 73x - 23
\][/tex]
4. Repeat the process with the new polynomial:
- Divide the leading terms: [tex]\(\frac{-44x^5}{3x^2} = -\frac{44}{3}x^3\)[/tex].
- Multiply the entire divisor by this result:
[tex]\[
-\frac{44}{3}x^3 \cdot (3x^2 + 7x + 11) = -44x^5 - \frac{308}{3}x^4 - \frac{484}{3}x^3
\][/tex]
- Subtract:
[tex]\[
(-44x^5 + x^4 + 70x^3 + 73x - 23) - (-44x^5 - \frac{308}{3}x^4 - \frac{484}{3}x^3) = \left(1 + \frac{308}{3}\right)x^4 + \left(70 + \frac{484}{3}\right)x^3 + 73x - 23
\][/tex]
[tex]\[
= \frac{311}{3}x^4 + \frac{694}{3}x^3 + 73x - 23
\][/tex]
5. Continue this process iteratively:
- Next division: [tex]\(\frac{\frac{311}{3}x^4}{3x^2} = \frac{311}{9}x^2\)[/tex].
- Multiply and subtract:
[tex]\[
\frac{311}{9}x^2 \cdot (3x^2 + 7x + 11) = \frac{311}{3}x^4 + \frac{2177}{9}x^3 + \frac{3421}{9}x^2
\][/tex]
[tex]\[
\left(\frac{311}{3}x^4 + \frac{694}{3}x^3 + 73x - 23\right) - \left(\frac{311}{3}x^4 + \frac{2177}{9}x^3 + \frac{3421}{9}x^2\right) = -\frac{9598}{27}x^2 + 73x - 23
\][/tex]
6. Proceed with next division and subtraction:
- Next division: [tex]\(\frac{-\frac{9598}{27}x^2}{3x^2} = -\frac{9598}{81}\)[/tex].
- Resulting polynomial becomes:
[tex]\[
-\frac{9598}{81}(3x^2 + 7x + 11) = -\frac{9598}{27}x^2 - \frac{67186}{81}x - \frac{105578}{81}
\][/tex]
[tex]\[
\left(-\frac{9598}{27}x^2 + 73x - 23\right) - \left(-\frac{9598}{27}x^2 - \frac{67186}{81}x - \frac{105578}{81}\right) = \left(73 + \frac{67186}{81}\right)x + \left(-23 + \frac{105578}{81}\right)
\][/tex]
[tex]\[
= \frac{76234}{81}x + \frac{103715}{81}
\][/tex]
Thus, our quotient is [tex]\(5x^4 - \frac{44}{3}x^3 + \frac{311}{9}x^2 - \frac{9598}{81}\)[/tex] and the remainder is [tex]\(\frac{76234}{81}x + \frac{103715}{81}\)[/tex].
In summarized form:
[tex]\[
\frac{15x^6 - 9x^5 + 56x^4 + 70x^3 + 73x - 23}{3x^2 + 7x + 11} = 5x^4 - \frac{44}{3}x^3 + \frac{311}{9}x^2 - \frac{95}{27}x - \frac{9598}{81} + \frac{\frac{76234}{81}x + \frac{103715}{81}}{3x^2 + 7x + 11}
\][/tex]
