Question

Factor f(x)=x4-x3-7x2+13x-6 completely. Then Sketch the graph

Answer 1

ANSWER TO QUESTION 1

$f(x)=(x-2)(x+3)(x-1)^2$.

EXPLANATION

The function given to us is,

$f(x)=x^4-x^3-7x^2+13x-6$

According to rational roots theorem,

$\pm1,\pm2,\pm3,\pm6$ are possible rational zeros of  

$f(x)=x^4-x^3-7x^2+13x-6$.

We find out that,

$f(-3)=(-3)^4-(-3)^3-7(-3)^2+13(-3)-6$

$f(-3)=81+27-63-39-6$

$f(-3)=6-6$

$f(-3)=0$

Also

$f(2)=(2)^4-(2)^3-7(2)^2+13(2)-6$

$f(2)=16-8-28+26-6$

$f(2)=6-6$

$f(2)=0$

This implies that  

$x-2 and x+3$ are factors of

$f(x)=x^4-x^3-7x^2+13x-6$ and hence $(x-2)(x+3)=x^2+x-6$ is also a factor.

We perform the long division as shown in the diagram.

Hence,

$f(x)=(x-2)(x+3)(x-1)^2$.

ANSWER TO QUESTION 2

Sketching the graph

We can see from the factorization that the roots  

$x=2$ and $x=-3$ have a multiplicity of 1, which is odd. This means that the graph crosses the x-axis at this intercepts.

Also the root $x=1$ has a multiplicity of 2, which is even. This means the graph does not cross the x-axis at this intercept.

Now we determine the position of the graph on the following intervals,

$x\le -3$

$f(-4)=(-4)^4-(-4)^3-7(-4)^2+13(-4)-6$

$f(-4)=150\:>0$

$-3\le x \le 1$

$f(0)=-6\:$

$1\le x\le 2$

$f(1.5)=-0.56\:$

$x \ge 2$

$f(3)=24\:>0$

We can now use these information to sketch the function as shown in diagram