ANSWER TO QUESTION 1
[tex]f(x)=(x-2)(x+3)(x-1)^2[/tex].
EXPLANATION
The function given to us is,
[tex]f(x)=x^4-x^3-7x^2+13x-6[/tex]
According to rational roots theorem,
[tex]\pm1,\pm2,\pm3,\pm6[/tex] are possible rational zeros of
[tex]f(x)=x^4-x^3-7x^2+13x-6[/tex].
We find out that,
[tex]f(-3)=(-3)^4-(-3)^3-7(-3)^2+13(-3)-6[/tex]
[tex]f(-3)=81+27-63-39-6[/tex]
[tex]f(-3)=6-6[/tex]
[tex]f(-3)=0[/tex]
Also
[tex]f(2)=(2)^4-(2)^3-7(2)^2+13(2)-6[/tex]
[tex]f(2)=16-8-28+26-6[/tex]
[tex]f(2)=6-6[/tex]
[tex]f(2)=0[/tex]
This implies that
[tex]x-2 and x+3[/tex] are factors of
[tex]f(x)=x^4-x^3-7x^2+13x-6[/tex] and hence [tex](x-2)(x+3)=x^2+x-6[/tex] is also a factor.
We perform the long division as shown in the diagram.
Hence,
[tex]f(x)=(x-2)(x+3)(x-1)^2[/tex].
ANSWER TO QUESTION 2
Sketching the graph
We can see from the factorization that the roots
[tex]x=2[/tex] and [tex]x=-3[/tex] have a multiplicity of 1, which is odd. This means that the graph crosses the x-axis at this intercepts.
Also the root [tex]x=1[/tex] 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,
[tex]x\le -3[/tex]
[tex]f(-4)=(-4)^4-(-4)^3-7(-4)^2+13(-4)-6[/tex]
[tex]f(-4)=150\:>0[/tex]
[tex]-3\le x \le 1[/tex]
[tex]f(0)=-6\:<0[/tex]
[tex]1\le x\le 2[/tex]
[tex]f(1.5)=-0.56\:<0[/tex]
[tex]x \ge 2[/tex]
[tex]f(3)=24\:>0[/tex]
We can now use these information to sketch the function as shown in diagram
