GIVEN:
We are given the following polynomial;
[tex]f(x)=x^4-2x^3-25x^2+2x+24[/tex]
Required;
We are required to sketch the graph of the function. Also, to use the synthetic division and the remainder theorem to find the zeros.
Step-by-step solution;
We shall begin by sketching a graph of the polynomial function.
From the graph of this polynomial, we can see that there are four points where the graph crosses the x-axis. These are the zeros of the function. One of the zeros is at the point;
[tex](-1,0)[/tex]
That is, where x = -1, and y = 0.
We shall take this factor and divide the polynomial by this factor.
The step by step procedure is shown below;
Now we have the coefficients of the quotient as follows;
[tex]1,-3,-22,24[/tex]
That means the quotient is;
[tex]x^3-3x^2-22x+24[/tex]
We can also divide this by (x - 1) and we'll have;
We now have the coefficients of the quotient after dividing a second time and these are;
[tex]x^2-2x-24[/tex]
The remaining two factors are the factors of the quadratic expression we just arrived at.
We can factorize this and we'll have;
[tex]\begin{gathered} x^2-2x-24 \\ \\ x^2+4x-6x-24 \\ \\ (x^2+4x)-(6x+24) \\ \\ x(x+4)-6(x+4) \\ \\ (x-6)(x+4) \end{gathered}[/tex]
The zeros of this polynomial therefore are;
[tex]\begin{gathered} f(x)=x^4-2x^3-25x^2+2x+24 \\ \\ f(x)=(x+1)(x-1)(x-6)(x+4) \\ \\ Where\text{ }f(x)=0: \\ \\ (x+1)(x-1)(x-6)(x+4)=0 \end{gathered}[/tex]
Therefore;
ANSWER:
[tex]\begin{gathered} x+1=0,\text{ }x=-1 \\ \\ x-1=0,\text{ }x=1 \\ \\ x-6=0,\text{ }x=6 \\ \\ x+4=0,\text{ }x=-4 \end{gathered}[/tex]
