Answer :
We start with the polynomial
[tex]$$
2x^4 - 13x^3 - 20x^2 - 39x + 56,
$$[/tex]
and we are given that [tex]$(x - 8)$[/tex] is one factor. To find the other factor, we can divide the polynomial by [tex]$(x - 8)$[/tex] using synthetic division.
1. Write the coefficients of the polynomial:
[tex]$$
2,\quad -13,\quad -20,\quad -39,\quad 56.
$$[/tex]
2. Since the factor is [tex]$(x - 8)$[/tex], we use [tex]$8$[/tex] for the synthetic division.
3. Set up the synthetic division:
[tex]\[
\begin{array}{r|rrrrr}
8 & 2 & -13 & -20 & -39 & 56 \\
& & 16 & 24 & 4 & -7 \\
\hline
& 2 & 3 & 4 & -7 & 0 \\
\end{array}
\][/tex]
- Bring the first coefficient down: [tex]$2$[/tex].
- Multiply [tex]$2$[/tex] by [tex]$8$[/tex] to get [tex]$16$[/tex], then add to [tex]$-13$[/tex] to get [tex]$3$[/tex].
- Multiply [tex]$3$[/tex] by [tex]$8$[/tex] to get [tex]$24$[/tex], then add to [tex]$-20$[/tex] to get [tex]$4$[/tex].
- Multiply [tex]$4$[/tex] by [tex]$8$[/tex] to get [tex]$32$[/tex], then add to [tex]$-39$[/tex] to get [tex]$-7$[/tex].
- Multiply [tex]$-7$[/tex] by [tex]$8$[/tex] to get [tex]$-56$[/tex], then add to [tex]$56$[/tex] to get [tex]$0$[/tex], which confirms [tex]$(x-8)$[/tex] is indeed a factor.
4. The resulting coefficients from the synthetic division are [tex]$2$[/tex], [tex]$3$[/tex], [tex]$4$[/tex], and [tex]$-7$[/tex]. This gives us the cubic factor
[tex]$$
2x^3 + 3x^2 + 4x - 7.
$$[/tex]
Thus, we have factored the original polynomial as
[tex]$$
2x^4 - 13x^3 - 20x^2 - 39x + 56 = (x - 8)\left(2x^3 + 3x^2 + 4x - 7\right).
$$[/tex]
The required factor is
[tex]$$
\boxed{2x^3 + 3x^2 + 4x - 7}.
$$[/tex]
[tex]$$
2x^4 - 13x^3 - 20x^2 - 39x + 56,
$$[/tex]
and we are given that [tex]$(x - 8)$[/tex] is one factor. To find the other factor, we can divide the polynomial by [tex]$(x - 8)$[/tex] using synthetic division.
1. Write the coefficients of the polynomial:
[tex]$$
2,\quad -13,\quad -20,\quad -39,\quad 56.
$$[/tex]
2. Since the factor is [tex]$(x - 8)$[/tex], we use [tex]$8$[/tex] for the synthetic division.
3. Set up the synthetic division:
[tex]\[
\begin{array}{r|rrrrr}
8 & 2 & -13 & -20 & -39 & 56 \\
& & 16 & 24 & 4 & -7 \\
\hline
& 2 & 3 & 4 & -7 & 0 \\
\end{array}
\][/tex]
- Bring the first coefficient down: [tex]$2$[/tex].
- Multiply [tex]$2$[/tex] by [tex]$8$[/tex] to get [tex]$16$[/tex], then add to [tex]$-13$[/tex] to get [tex]$3$[/tex].
- Multiply [tex]$3$[/tex] by [tex]$8$[/tex] to get [tex]$24$[/tex], then add to [tex]$-20$[/tex] to get [tex]$4$[/tex].
- Multiply [tex]$4$[/tex] by [tex]$8$[/tex] to get [tex]$32$[/tex], then add to [tex]$-39$[/tex] to get [tex]$-7$[/tex].
- Multiply [tex]$-7$[/tex] by [tex]$8$[/tex] to get [tex]$-56$[/tex], then add to [tex]$56$[/tex] to get [tex]$0$[/tex], which confirms [tex]$(x-8)$[/tex] is indeed a factor.
4. The resulting coefficients from the synthetic division are [tex]$2$[/tex], [tex]$3$[/tex], [tex]$4$[/tex], and [tex]$-7$[/tex]. This gives us the cubic factor
[tex]$$
2x^3 + 3x^2 + 4x - 7.
$$[/tex]
Thus, we have factored the original polynomial as
[tex]$$
2x^4 - 13x^3 - 20x^2 - 39x + 56 = (x - 8)\left(2x^3 + 3x^2 + 4x - 7\right).
$$[/tex]
The required factor is
[tex]$$
\boxed{2x^3 + 3x^2 + 4x - 7}.
$$[/tex]
