Answer
The roots of the polynomial are\[\left\{ -4,-2,-\frac{1}{2},\frac{1}{2},3 \right\}\]
Work Step by Step
Consider the given polynomial $4{{x}^{5}}+12{{x}^{4}}-41{{x}^{3}}-99{{x}^{2}}+10x+24=0.$.
Determine values of p and q where p are the factors of the constant in the polynomial and q are the factors of the leading coefficient of the polynomial.
$p=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24$
$q=\pm 1,\pm 2,\pm 4$
Calculate $\frac{p}{q}$.
$\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm 12,\pm 24,\pm \frac{1}{2},\pm 6,\pm \frac{1}{4},\pm 3,\pm 2$
According to Descartes’s rule of signs, the function $4{{x}^{5}}+12{{x}^{4}}-41{{x}^{3}}-99{{x}^{2}}+10x+24=0$ has two sign changes. Since the function $f\left( x \right)$ has two variations in sign, the function $f\left( x \right)$ has two or zero positive real root.
Further, evaluate $f\left( -x \right):$
$-4{{x}^{5}}+12{{x}^{4}}+41{{x}^{3}}-99{{x}^{2}}-10x+24=0$
Since the function $f\left( x \right)$ has three variations in sign, the function $f\left( x \right)$ has three or one negative real root.
Obtain the roots of the provided polynomial, graph the function and obtain the value of x where the graph intersects the x-axis.
Observe from the graph that the roots of the provided equation are $-4,-2,-0.5,0.5,3$.
Therefore, the roots of the provided function are $\left\{ -4,-2,-\frac{1}{2},\frac{1}{2},3 \right\}$.
