Can we factorise $6x^3+5x^2-17x-6$ completely?

Can we factorise $6x^3+5x^2-17x-6$ completely?

Question

1. Given that $f(x)=6x^3+5x^2-17x-6$, find the remainder when $f(x)$ is divided by $x-2$, find the remainder when $f(x)$ is divided by $x+2$, factorise $f(x)$ completely.

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1. Given that the expression $ax^3-x^2+bx+18$ is exactly divisible by $x^2+x-6$, find the value of $a$ and $b$.

Suggestion

Would it help us to recall the remainder theorem?

$\qquad$ iii. factorise $f(x)$ completely.

Can we find a remainder that is zero? What does this tell us about the roots of $f(x)$?

Would it help to try to factorise $x^2+x-6$?

Solution

The remainder theorem tells us that if $f(c)=r$ for some polynomial $f(x)$, then $r$ is the remainder when $f(x)$ is divided by $x-c$.

In our case, $c$ is $2$ for the first part, and $-2$ for the second.

We can now substitute these values into $f(x)$, giving

$$\begin{align*} f(+2)&=28,\\ f(-2)&=0. \end{align*}$$

Therefore the remainder when $f(x)$ is divided by $x-2$ is $28$, and the remainder on dividing by $x+2$ is $0$.

$\qquad$ (iii) factorise $f(x)$ completely.

Since the remainder of $f(x)$ divided by $x+2$ is zero, we know $x+2$ is a factor of $f(x)$.

Check the factor theorem if you need to…

We now know that $f(x)$ is of the form $(x+2)(ax^2+bx+c)$, where $a$, $b$ and $c$ are constants to be found. This means $f(x) = 6x^3 + 5x^2 - 17x - 6 = (x+2)(ax^2+bx+c).$

We can find $a$, $b$ and $c$ by considering the coefficients of the different powers of $x$.

The only cubic term on the right hand side is $ax^3$, so we can see by comparing it with the left hand side that $a = 6$.

The constant term is $2c$, which must equal $-6$, and so $c = -3$.

There’ll be two square terms in the expansion of the right hand side, namely

$12x^2$

and

$bx^2$

. Therefore

$$\begin{align*} 12 + b &= 5 \\ \implies b &= -7. \end{align*}$$

We can check this by looking at the linear terms; on the right hand side, we get

$2bx$

and

$-3x$

, so

$$\begin{align*} 2b - 3 &= -17 \\ \implies 2b &= -14 \\ \implies b &= -7, \quad \text{as above}. \end{align*}$$

Therefore we can write $f(x) = (x + 2)(6x^2 - 7x - 3)$.

The quadratic bracket here can be factorised by inspection, or by using the quadratic formula if we’re stuck. We find that $f(x) = (x + 2)(3x + 1)(2x - 3).$

Or we could use polynomial division…

We could also have found the quadratic $6x^2 - 7x - 3$ by using polynomial division. $\begin{array}{r} 6x^2\phantom{1}{}-7x-3 x+2 {6x^3\phantom{1}{}+5x^2-17x-6} \underline{6x^3+12x^2}\phantom{{}-17x-6} \phantom{6x^3}{}-7x^2-17x\phantom{{}-6} \phantom{6x^3}{}\underline{{}-7x^2-14x}\phantom{{}-6} \phantom{6x^3+5x^2}{}-3x-6 \phantom{6x^3+5x^2}\underline{{}-3x-6} \phantom{6x^3+5x^2-17x+{}}0 \end{array}$

Although the question doesn’t ask for this, click below for a plot of $y=f(x)$, with the three roots indicated.

Show Plot

Let’s factorise; $x^2 + x - 6 = (x + 3)(x - 2)$.

So we know that $(x-2)$ and $(x+3)$ are factors of the above cubic expression, and so if we substitute in $x=2$ or $x=-3$ to this cubic, the result has to be zero.

This leads to the following pair of equations:

$$\begin{align} 8a+2b+14&=0 \quad \implies \quad 4a + b + 7 = 0 \\ -27a-3b+9&=0 \quad \implies \quad -9a - b + 3 = 0. \end{align}$$

We can solve these by adding them together:

$$\begin{align*} -5a + 10 &= 0 \\ \implies a &= 2, \end{align*}$$

and substituting into

$(I)$

, we find

$$\begin{align*} 4 \times 2 + b + 7 &= 0 \\ \implies b &= -15. \end{align*}$$

We can check this by substituting both $a$ and $b$ into $(II)$: $-9 \times 2 - (-15) + 3 = 0.$

We therefore find $a= 2$, $b = 15$, and that the full cubic function is $2x^3-x^2-15x+18$.