How to factor out this expression

Question

I'm practicing limits and I have to factor out $(x-3)$ from the following expression: $$2x^3-19x+3$$

Answer 1

Using synthetic division, we have that . $$\begin{array}{c|rrr} & 2 & 0 & -19 & 3\\ {\color{red}3} & \downarrow & 6 & 18 & -3\\ \hline & 2 & 6 & -1 & |\phantom{-} {\color{blue}0} \end{array}$$ Thus, $$2x^3-19x+3=(x-3)(2x^2+6x-1)$$

Answer 2

You can factor as follows:

$$2x^3-19x+3=2x^3-6x^2+6x^2-18x-x+3=$$ $$=2x^2(x-3)+6x(x-3)-(x-3)=$$ $$=(x-3)(2x^2+6x-1)$$

Answer 3

Assuming you never heard of long division, you can guess that the first term of the ratio $\dfrac{2x^3-19x+3}{x-3}$ will be $2x^2$, because when $x$ grows, the lower powers become neglectible, and the division "simplifies" to $\dfrac{2x^3}x$.

Then

$$2x^3-19x+3=(x-3)(2x^2+R(x))$$ where $R(x)$ is some remainder, a polynomial of the first degree.

Now expanding and simplifying,

$$2x^3-19x+3-(x-3)2x^2=6x^2-19x+3=(x-3)R(x).$$

So you have to perform the division $\dfrac{6x^2-19x+3}{x-3}$.

Similarly, the next term will be $6x$, and

$$6x^2-19x+3-(x-3)6x=-x+3=(x-3)R'(x).$$

You should be able to conclude and collect the results.

Answer 4

Long division: $$ \begin{array}{rrrrrrrrrrrrrrr} & & & 2x^2 & +6x & -1\\ & & --- & --- & --- & --- \\ 1x-3 & ) & 2x^3 & 0x^2 & -19x & +3 \\ & & 2x^3 & -6x^2 \\ & & --- & --- \\ & & & 6x^2 & -19x \\ & & & 6x^2 & -18x \\ & & & --- & --- \\ & & & & -1x & +3 \\ & & & & -1x & +3 \\ & & & & --- & --- \\ & & & & & 0 \\ \end{array} $$

Answer 5

Undetermined coefficients is easier than long division here. Comparing (coef's of) $\rm\color{#0a0}{highest}$ and $\color{#c00}{\rm lowest}$ degree terms on both sides yields the coef's of highest and lowest terms of the quadratic

$$ \color{#0a0}{2x^3}-19x+\color{#c00}3\ =\, (\color{#0a0}x\color{#c00}{-3})(\color{#0a0}{2x^2}+bx\color{#c00}{-1})$$

Comparing coefficients of $x^2$ yields $\, 0 = b-6,\ $ so $\ b=6$