Question

question 3 given the function $p(n)=4n^{4}-40n^{3}+84n^{2}$ its $p$-intercept is its $n$-intercepts are question help: video message instructor submit question question 4 use your graphing calculator to solve the equation graphically for all real solutions $x^{3}-5x^{2}+2x + 12 = 0$ solutions: $x=$ make sure your answers are accurate to at least two decimals question help: video message instructor

Explanation:

Step1: Find the $P$-intercept

Set $n = 0$ in $P(n)=4n^{4}-40n^{3}+84n^{2}$. Then $P(0)=4\times0^{4}-40\times0^{3}+84\times0^{2}=0$.

Step2: Find the $n$-intercepts

Set $P(n)=0$, so $4n^{4}-40n^{3}+84n^{2}=0$. Factor out $4n^{2}$: $4n^{2}(n^{2}-10n + 21)=0$.
Factor the quadratic: $4n^{2}(n - 3)(n - 7)=0$.
Set each factor equal to zero:

• For $4n^{2}=0$, $n = 0$.
• For $n - 3=0$, $n=3$.
• For $n - 7=0$, $n = 7$.

Step3: Solve $x^{3}-5x^{2}+2x + 12=0$ (using rational - root theorem and synthetic division or graphing)

By the rational - root theorem, the possible rational roots are factors of $12$, i.e., $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$.
Testing $x=-1$: $(-1)^{3}-5(-1)^{2}+2(-1)+12=-1 - 5-2 + 12=4
eq0$.
Testing $x = 2$: $2^{3}-5\times2^{2}+2\times2+12=8-20 + 4+12=4
eq0$.
Testing $x=3$: $3^{3}-5\times3^{2}+2\times3+12=27-45 + 6+12=0$.
Since $x = 3$ is a root, $(x - 3)$ is a factor. Using synthetic division:

3 |  1  -5  2  12
   |     3 -6 -12
   |  1  -2 -4   0

The quotient is $x^{2}-2x - 4$.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $x^{2}-2x - 4$ where $a = 1$, $b=-2$, $c=-4$.
$x=\frac{2\pm\sqrt{(-2)^{2}-4\times1\times(-4)}}{2\times1}=\frac{2\pm\sqrt{4 + 16}}{2}=\frac{2\pm\sqrt{20}}{2}=\frac{2\pm2\sqrt{5}}{2}=1\pm\sqrt{5}$.
$x=1+\sqrt{5}\approx1 + 2.24=3.24$ and $x=1-\sqrt{5}\approx1-2.24=-1.24$.

Answer:

$P$-intercept: $0$
$n$-intercepts: $0,3,7$
Solutions for $x^{3}-5x^{2}+2x + 12=0$: $x = 3,x\approx-1.24,x\approx3.24$