Question

One root of f(x) = 2x3 9x2 7x – 6 is –3. explain how to find the factors of the polynomial.

Answer 1

Identify the factor x + 3 from the given root –3.

Use synthetic division to divide the polynomial by x + 3.

Use the bottom row of the synthetic division as coefficients in the quadratic 2x2 + 3x – 2.

Factor the quotient to find the two other factors: x + 2 and 2x – 1.

Answer 2

We have the following polynomial:

$f(x)=2x^{3}+9x^{2}+7x-6$

The problem states that one root is -3. Thus, it is true that $(x+3)$ is a factor of the polynomial. Given that this is fulfilled, it is also true that:

$f(x)=(x+3)Q(x) \therefore Q(x)=\frac{f(x)}{x+3} \\ \\ where \ Q(x) \ has \ a \ degree \ of \ 2$

We can find Q(x) by applying Ruffini's rule, thus:

$\ \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ 7 \ \ \ \ \ \ -6 \\ -3 \\ \rule{50mm}{0.1mm} \\ \ {} \ \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ 3 \ \ \ \ \ -2 \ \ \ \ \ \ \ \ 0$

Therefore:

$Q(x)=2x^{2}+3x-2$

The roots of this polynomial can be get as follows:

$x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \rightarrow x_{12}=\frac{-3\pm\sqrt{3^2-4(2)(-2)}}{2(2)}\\x_{1}=\frac{1}{2};\ x_{2}=-2$

These are the roots along with $-3$. Finally, the factored polynomial can be written as follows:

$f(x)=(x+3)(x+2)(2x-1)$