Question

Use the Factor Theorem and synthetic division to show x + 5 is a factor of f(x) = 2x3 + 7x2 − 14x + 5

Answer 1

Answer:

Factor theorem: $f(-5) = 0$.

Synthetic division: $f(x) = (x + 5)\, (2\, x^2 -3\, x + 1)$.

Step-by-step explanation:

Factor Theorem

By the factor theorem, a monomial of the form $(x - a)$ (where $a$ is a constant) is a factor of polynomial $f(x)$ if and only if $f(a) = 0$.

In this question, the monomial is $(x + 5)$, which is equivalently $(x - (-5))$. $a = -5$.

$\begin{aligned}& f(-5) \\ &= 2 \times (-5)^{3} + 7 \times (-5)^{2}- 14\times (-5) + 5\\ &= -250 + 175 + 70 + 5 \\ &= 0\end{aligned}$.

Hence, by the factor theorem,  $(x + 5)$, which is equivalent to $(x - (-5))$, is a factor of $f(x)$.

Synthetic Division

$\begin{aligned}& f(x) \\ &= 2\, x^{3} + 7\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} - 3\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} + \underbrace{(x + 5)\, (-3\, x)}_{-3\, x^{2} - 15\, x} + (x + 5) \\ &= (x + 5)\, (2\, x^{2} - 3\, x + 1)\end{aligned}$.

The remainder is $0$ when dividing $f(x)$ by $(x + 5)$. Hence, $(x + 5)\!$ is a factor of $f(x)\!$.