Show that $1$ and $2$ are zeros of the following polynomial

Question

Show that $1$ and $2$ are zeros of the polynomial $P(x)=x^4-2x^3+5x^2-16x+12$ and hence that $(x-1)(x-2)$ is a factor of $P(x)$

Answer 1

A zero of a polynomial is a value of $x$ for which $P(x) = 0$.

To show that $1 \text{ and}\; 2$ are zeros, substitute each value into the polynomial $P(x)$ and evaluate the polynomial to confirm that it evaluates to $0$:

Determine if $P(1) = 0? $ and if $P(2) = 0?$...

That is, what do you get when you let $\;x = 1?\;$ What do you get when you let $\;x = 2$?
If $P(x) = 0$ when $x = 1$, then $x = 1$ is a zero of $P(x)$. And if $P(x) = 0 $ when $x = 2$, then $x = 2$ is a zero of $P(x)$

Note that if $x_0$ is a zero of a polynomial, then $(x - x_0)$ is a factor of the polynomial.
So once you confirm that $P(1) = P(2) = 0$, you will know that $(x - 1)(x-2)$ is a factor of $P(x)$: when either factor evaluates to zero, the entire factored function $P(x)$ evaluates to zero.

Good to know: whenever you are given a factored polynomial - suppose you are given that $Q(x) = (x-1)(x-2)(x-3)$ - you can easily "read off" its "zeros".

Recall that anything multiplied by zero is equal to zero.:

So when will $\;\bf Q(x) = (x-1)(x-2)(x-3)= 0\;?$

• $Q(x) = 0\;$ when $\;(x - 1) = 0\implies x = 1$,

• $Q(x) = 0\;$ when $\;(x - 2) = 0 \implies x = 2$,

• $Q(x) = 0\;$ when $\;(x - 3) = 0 \implies x = 3$.

  Any of those values for $\,x: x= 1, \,x = 2,\,\text{ or}\; x= 3,\,$ will give you $Q(x) = 0$.

Answer 2

Recall that $(x-a)$ is a factor of a polynomial $P(x)$, if $P(a) = 0$. Hence, to check if $(x-1)$ and $(x-2)$ are factors, find the value of $P(1)$ and $P(2)$ to conclude what you want.

Answer 3

You just have to substitute x by 1 and by 2. This will give you $P(1)=0$ and $P(2)=0$ is these values are eventually zeros. If they are, you could factorize them by $(x-1)(x-2)$ because the zeros of this expression are 1 and 2.

Answer 4

The is an application of the Factor Theorem.

If $\operatorname{f}(x)$ is a polynomial and $\operatorname{f}(p)=0$, then $x-p$ is a factor of $\operatorname{f}(x)$. In particular: if $\operatorname{f}(1)=0$ then $x-1$ is a factor of $\operatorname{f}(x)$ and if $\operatorname{f}(2)=0$ then $x-2$ is a factor of $\operatorname{f}(x)$. Hence to show that $\operatorname{f}(x)$ has $(x-1)(x-2)$ as a factor, we must show that both $\operatorname{f}(1)=0$ and $\operatorname{f}(2)=0$. Notice that:

\begin{array}{ccccc} \operatorname{f}(1) &=& 1^4 - 2 \times 1^3 + 5 \times 1^2 - 16 \times 1 +12 &=& 1 - 2 + 5 - 16 + 12 &=& 0 \\ \operatorname{f}(2) &=& 2^4 - 2 \times 2^3 + 5 \times 2^2 - 16 \times 2 +12 &=& 16 - 16 + 20 - 32 + 12 &=& 0 \end{array}

It follows that both $x-1$ and $x-2$ divide $\operatorname{f}(x)$ and so their product $(x-1)(x-2)$ divides $\operatorname{f}(x)$.