Algebra for Home Schooling

Question

Check whether $(x+1)$ is a factor of the polynomial $x⁴+x³+x²-5x+1$ .Justify your answer? What I did was:

$$\begin{align*}(x+1)&=0 \\\implies x &= -1 \end{align*}$$

Then, $x^4+x^3+x^2-5x+1= (-1^4)+(-1^3)+(-1^2)-(-5)+1=7$

The answers should be $0$, shouldn't it?

Answer 1

Hint:

If $(x+1)$ is a factor of $x^4+x^3+x^2-5x+1$, then you can write $$x^4+x^3+x^2-5x+1=(x+1)g(x)$$ where $g(x)$ is a polynomial of degree $3$. Now, what happens if you let $x=-1$? (Compare RHS and LHS)

Answer 2

Hint: Using synthetic division with "$R$" value $-1$:

 -1 |  1   1   1   -5    1
 ‾‾‾      -1   0   -5   10
       1   0   1  -10   11

The remainder is not equal to $0$. What can you tell from that?

Answer 3

There are good answers here. Here's how to state your correct argument.

I know that $r$ is a root if and only if $x-r$ is a factor. When I evaluate the polynomial at $x=-1$ I get $7 \ne 0$ so $x+1$ is not a factor.

Your math was right but your arithmetic was wrong: the value of the polynomial at $x=-1$ is $7$, not $1$.

If you ask more questions here, show your work and use mathjax: MathJax basic tutorial and quick reference

Answer 4

Firstly, people on here can be a bit pedantic so it will always serve you better to show your attempt. It really doesn't have to be right. It's just so that they know that they're not doing your work FOR you and you are at least aiming to understand the work not just copy.

Now for the Maths:
If $(x+1)$ is a factor of $f(x)=x^4+x^3+x^2−5x+1$ then $ x^4+x^3+x^2−5x+1=(x+1)g(x) : g(x)$ is some cubic function.
$\Rightarrow$ when $f(x)=0,$ $x^4+x^3+x^2−5x+1=(x+1)g(x)=0 \Rightarrow g(x)=0$ and $(x+1)=0 \Rightarrow x=-1$
In other words $f(x)=0$ when $x=-1$ but when you substitute $x=-1$ into $f(x)$, you get $f(-1)=(-1)^4+(-1)^3+(-1)^2-5(-1)+1=1-1+1+5+1=7\ne0 \Rightarrow -1$ is not a root $\Rightarrow (x+1)$ is not a factor.

Answer 5

If $(x+1)$ is a factor of the polynomial $p(x)$ then

$p(x)= (x+1)q(x)$ where $q(x)$ is a polynomial of one degree less.

Then:

$p(-1)=0.$

Since $p(-1) \not =0$ in our case, $(x+1)$ is not a factor .

Answer 6

$$x^4+x^3+x^2-5x+1$$

A good way to check it is to factorize it to see if $x+1$ is the factor of the polynomial:

$x^4+x^3+x^2-5x+1$

$=x^4+x^3+x^2+x-6x-6+7$

$=x^3(x+1)+x(x+1)-6(x+1)+7$

$=(x+1)(x^3+x-6)+7$

We can conclude that:

$x+1$ is not a factor of the expression

and

$x^4+x^3+x^2-5x+1$ divided by $x+1$ will have $7$ as the remainder.