Solution:
Given:
[tex]f(x)=x^4+10x^3-15x^2-40x+44[/tex]
The zeros of the function are as follows;
Using the theorem, if f(a) = 0, then x = a is a root or zero.
Hence;
[tex]\begin{gathered} f(1)=1^4+10(1^3)-15(1^2)-40(1)+44 \\ f(1)=1+10-15-41+44 \\ f(1)=0 \\ \\ Hence,\text{ } \\ x=1\text{ is a zero} \end{gathered}[/tex][tex]\begin{gathered} f(2)=2^4+10(2^3)-15(2^2)-40(2)+44 \\ f(2)=16+80-60-80+44 \\ f(2)=0 \\ \\ Hence,\text{ } \\ x=2\text{ is a zero} \end{gathered}[/tex]
Trying other numbers, it follows also that;
[tex]\begin{gathered} f(-2)=(-2)^4+10(-2^3)-15(-2^2)-40(-2)+44 \\ f(-2)=16-80-60+80+44 \\ f(-2)=0 \\ \\ Hence,\text{ } \\ x=-2\text{ is a zero} \end{gathered}[/tex]
The last root is;
[tex]\begin{gathered} f(-11)=(-11)^4+10(-11^3)-15(-11^2)-40(-11)+44 \\ f(-11)=14641-13310-1815+440+44 \\ f(-11)=0 \\ \\ Hence,\text{ } \\ x=-11\text{ is a zero} \end{gathered}[/tex]
Therefore, the zeros are;
x = 1
x = 2
x = -2
x = -11
Using the factor theorem, if x = a is a zero, then (x-a) is a factor.
Thus,
[tex]\begin{gathered} x=1,(x-1)\text{ is a factor} \\ x=2,(x-2)\text{ is a factor} \\ x=-2,(x+2)\text{ is a factor} \\ x=-11,(x+11)\text{ is a factor} \end{gathered}[/tex]
Hence, the polynomial can be factored as;
[tex]x^4+10x^3-15x^2-40x+44=\left(x-1\right)\left(x+11\right)\left(x+2\right)\left(x-2\right)[/tex]
