# How do you find all rational zeroes of the function using synthetic division f(x)=x^3-7x^2+2x+40?

##### 1 Answer
Oct 6, 2016

All the rational zeroes of $f \left(x\right) = 0$are, $- 2 , 4 , \mathmr{and} , 5.$

#### Explanation:

We can easily see that $\left(x \pm 1\right)$ are not the factors of $f \left(x\right) .$

Now, let us observe that, the leading co-eff. of $f \left(x\right)$ is $1$, and has

factor $1$, and, the const. term is $40$, having factors,

$1 , 2 , 4 , 5 , 8 , 10 , 20 , 40.$

Therefore, by the Rational Root Theorem, the probable factors can

be worked out :-

$1 x \pm 1 , 1 x \pm 2 , 1 x \pm 4 , x \pm 5 , x \pm 8 , x \pm 10 , \pm 20 , x \pm 40$

Of these, $x \pm 1$ have already been considered as non-factors.

As for, $x - 2$, we have, $f \left(2\right) = 8 - 28 + 4 + 40 \ne 0.$ It is not a factor.

For, $x + 2 , f \left(- 2\right) = - 8 - 28 - 4 + 40 = 0 \therefore \left(x + 2\right)$ is a factor.

Now to factorise $f \left(x\right)$ completely, the Long / Synthetic Division can

be used. Instead, we proceed as under :-

$f \left(x\right) = {x}^{3} - 7 {x}^{2} + 2 x + 40$

$= \underline{{x}^{3} + 2 {x}^{2}} - \underline{9 {x}^{2} - 18 x} + \underline{20 x + 40}$

$= {x}^{2} \left(x + 2\right) - 9 x \left(x + 2\right) + 20 \left(x + 2\right)$

$= \left(x + 2\right) \left({x}^{2} - 9 x + 20\right)$

$= \left(x + 2\right) \left\{\underline{{x}^{2} - 5 x} - \underline{4 x + 20}\right\} \ldots \ldots . . \left[5 \times 4 = 20 , 5 + 4 = 9\right]$

$= \left(x + 2\right) \left\{x \left(x - 5\right) - 4 \left(x - 5\right)\right\}$

$= \left(x + 2\right) \left(x - 5\right) \left(x - 4\right) .$

Thus, all the rational zeroes of $f \left(x\right) = 0$are, $- 2 , 4 , \mathmr{and} , 5.$