How do you find the zeroes of P(x)= x^4-2x^3-7x^2+8x+12?

Mar 17, 2018

Find zeros: $x = - 1$, $x = 2$, $x = - 2$ and $x = 3$

Explanation:

Given:

$P \left(x\right) = {x}^{4} - 2 {x}^{3} - 7 {x}^{2} + 8 x + 12$

Rational roots theorem

The rational roots theorem tells us that the rational zeros of $P \left(x\right)$ are expressible as $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $12$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$

Trying each in turn, we find:

$P \left(- 1\right) = 1 + 2 - 7 - 8 + 12 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor.

We could carry on just trying other possible rational roots from the list, but let's factor out the one we have found...

${x}^{4} - 2 {x}^{3} - 7 {x}^{2} + 8 x + 12 = \left(x + 1\right) \left({x}^{3} - 3 {x}^{2} - 4 x + 12\right)$

Notice that in the remaining cubic factor the ratio of the first and second terms is the same as the ratio of the third and fourth terms.

So this cubic will factor by grouping:

${x}^{3} - 3 {x}^{2} - 4 x + 12 = \left({x}^{3} - 3 {x}^{2}\right) - \left(4 x - 12\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} - 4 x + 12} = {x}^{2} \left(x - 3\right) - 4 \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} - 4 x + 12} = \left({x}^{2} - 4\right) \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} - 4 x + 12} = \left({x}^{2} - {2}^{2}\right) \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} - 4 x + 12} = \left(x - 2\right) \left(x + 2\right) \left(x - 3\right)$

So the other three zeros are $\pm 2$ and $3$.

Mar 17, 2018

Four zeros are $- 1 , - 2 , 2$ and $3$.

Explanation:

According to Factor Theorem if for $f \left(x\right)$ a polynomial is divisible by $\left(x - a\right)$, then $f \left(a\right) = 0$. This makes $x = a$ a zero or root of a polynomial $f \left(x\right)$.

Further if $x = a$ is a polynomial with integral coeficients, with coefficient of highest degree being $1$, then $a$ is a factor of constant term.

Hence, here if $a$ is a zero of $P \left(x\right) = {x}^{4} - 2 {x}^{3} - 7 {x}^{2} + 8 x + 12$, it will be a factor of $12$ i.e. it can be among $\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$. Observe that product of zeros would be $12$, as we can write $P \left(x\right) = \left(x - {a}_{1}\right) \left(x - {a}_{2}\right) \left(x - {a}_{3}\right) \left(x - {a}_{4}\right)$ and then constant term is ${a}_{1} {a}_{2} {a}_{3} {a}_{4}$.

It is easier if $a = \pm 1$. Observe that if $a = 1$, $P \left(x\right) = 1 - 2 - 7 + 8 + 12 = 12$ and if $a = - 1$, $P \left(x\right) = 1 + 2 - 7 - 8 + 12 = 0$, hence while $- 1$ is a zero, $1$ is not a zero.

If $a = 2$ then $P \left(x\right) = 1 \cdot {2}^{4} - 2 \cdot {2}^{3} - 7 \cdot {2}^{2} + 8 \cdot 2 + 12 = 16 - 16 - 28 + 16 + 12 = 0$

and if $a = - 2$, $P \left(x\right) = 1 \cdot {\left(- 2\right)}^{4} - 2 \cdot {\left(- 2\right)}^{3} - 7 \cdot {\left(- 2\right)}^{2} + 8 \cdot \left(- 2\right) + 12 = 16 + 16 - 28 - 16 + 12 = 0$

Therefore, three zeros are $- 1 , 2$ and $- 2$ andfourth zero would be $\frac{12}{- 1 \cdot 2 \cdot - 2} = 3$. Let us check it.

If $a = 3$ then $P \left(x\right) = 1 \cdot {3}^{4} - 2 \cdot {3}^{3} - 7 \cdot {3}^{2} + 8 \cdot 3 + 12 = 81 - 54 - 63 + 24 + 12 = 0$

Mar 17, 2018

$- 2 , - 1 , 2 , 3$.

Explanation:

If we complete the square of ${x}^{4} - 2 {x}^{3}$, the last term

will be ${x}^{2}$.

Rearranging the terms of $P \left(x\right)$, we get,

$P \left(x\right) = \textcolor{g r e e n}{{x}^{4} - 2 {x}^{3} +} \textcolor{red}{{x}^{2} - {x}^{2}} \textcolor{g r e e n}{- 7 {x}^{2} + 8 x + 12}$,

$= \left({x}^{4} - 2 {x}^{3} + {x}^{2}\right) - \underline{8 {x}^{2} + 8 x} + 12$,

$= {\left({x}^{2} - x\right)}^{2} - 8 \left({x}^{2} - x\right) + 12$,

$= {y}^{2} - 8 y + 12 , \text{ say, where, } y = {x}^{2} - x$,

$= \underline{{y}^{2} - 6 y} - \underline{2 y + 12}$,

$= y \left(y - 6\right) - 2 \left(y - 6\right)$,

$= \left(y - 6\right) \left(y - 2\right)$,

$= \left\{\left({x}^{2} - x\right) - 6\right\} \left\{\left({x}^{2} - x\right) - 2\right\} \ldots \ldots \ldots . \left[\because , y = {x}^{2} - x\right]$,

$= \left\{\underline{{x}^{2} - 3 x} + \underline{2 x - 6}\right\} \left\{\underline{{x}^{2} - 2 x} + \underline{x - 2}\right\}$,

$= \left\{x \left(x - 3\right) + 2 \left(x - 3\right)\right\} \left\{x \left(x - 2\right) + 1 \left(x - 2\right)\right\}$,

$\Rightarrow P \left(x\right) = \left(x - 3\right) \left(x + 2\right) \left(x - 2\right) \left(x + 1\right)$.

Hence, the zeroes of $P \left(x\right)$ are, $- 2 , - 1 , 2 , 3$.