Product of a linear factor and cubic factor
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To follow up:
Since [MATH]P(-1)=0[/MATH] , we know by the factor theorem then that \(P(x)\) must have as a factor \(x+1\). We can use synthetic division to complete part (b):
[MATH]\begin{array}{c|rr}& 2 & -5 & 0 & 6 & -1 \\ -1 & & -2 & 7 & -7 & 1 \\ \hline & 2 & -7 & 7 & -1 & 0 \end{array}[/MATH]
And so, we conclude:
[MATH]P(x)=(x+1)\left(2x^3-7x^2+7x-1\right)[/MATH]
Since [MATH]P(-1)=0[/MATH] , we know by the factor theorem then that \(P(x)\) must have as a factor \(x+1\). We can use synthetic division to complete part (b):
[MATH]\begin{array}{c|rr}& 2 & -5 & 0 & 6 & -1 \\ -1 & & -2 & 7 & -7 & 1 \\ \hline & 2 & -7 & 7 & -1 & 0 \end{array}[/MATH]
And so, we conclude:
[MATH]P(x)=(x+1)\left(2x^3-7x^2+7x-1\right)[/MATH]