Question

Find the real zeros of F then use the real zeros to factor f.

Answer 1

Given:

`f(x)=x^4+10x^3-15x^2-40x+44`

The zeros of the function are as follows;

Using the theorem, if f(a) = 0, then x = a is a root or zero.

Hence;

`\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}`
`\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}`

Trying other numbers, it follows also that;

`\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}`

The last root is;

`\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}`

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,

`\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}`

Hence, the polynomial can be factored as;

`x^4+10x^3-15x^2-40x+44=\left(x-1\right)\left(x+11\right)\left(x+2\right)\left(x-2\right)`