Answer:
Factor theorem: [tex]f(-5) = 0[/tex].
Synthetic division: [tex]f(x) = (x + 5)\, (2\, x^2 -3\, x + 1)[/tex].
Step-by-step explanation:
Factor Theorem
By the factor theorem, a monomial of the form [tex](x - a)[/tex] (where [tex]a[/tex] is a constant) is a factor of polynomial [tex]f(x)[/tex] if and only if [tex]f(a) = 0[/tex].
In this question, the monomial is [tex](x + 5)[/tex], which is equivalently [tex](x - (-5))[/tex]. [tex]a = -5[/tex].
[tex]\begin{aligned}& f(-5) \\ &= 2 \times (-5)^{3} + 7 \times (-5)^{2}- 14\times (-5) + 5\\ &= -250 + 175 + 70 + 5 \\ &= 0\end{aligned}[/tex].
Hence, by the factor theorem, [tex](x + 5)[/tex], which is equivalent to [tex](x - (-5))[/tex], is a factor of [tex]f(x)[/tex].
Synthetic Division
[tex]\begin{aligned}& f(x) \\ &= 2\, x^{3} + 7\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} - 3\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} + \underbrace{(x + 5)\, (-3\, x)}_{-3\, x^{2} - 15\, x} + (x + 5) \\ &= (x + 5)\, (2\, x^{2} - 3\, x + 1)\end{aligned}[/tex].
The remainder is [tex]0[/tex] when dividing [tex]f(x)[/tex] by [tex](x + 5)[/tex]. Hence, [tex](x + 5)\![/tex] is a factor of [tex]f(x)\![/tex].
