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Part 1: Calculator Allowed

For questions 1-3, let $f\left(x\right)=3x^4-x^3-9x^2+3x$.

The graph is decreasing on $x\in \left(-\infty ,-1.1887\right)\cup \left(0.1651,1.2736\right)$ and increasing on $x\in \left(-1.1887,0.1651\right)\cup \left(1.2736,\infty \right)$.

There are minima at $x=-1.1887$ and $x=1.2736$. There is a maximum at $x=0.1651$.

I won’t re-graph or re-label…but they are at -1.7321, 0, 0.3333 and 1.7321.

$\begin{array}{rrrrrrr}\hfill & \hfill x^2& \hfill +2x& \hfill -6& \hfill & \hfill & \hfill \\ \hfill x^3+1& \hfill x^5& \hfill +2x^4& \hfill -6x^3& \hfill +x^2& \hfill -5x& \hfill +1\\ \hfill & \hfill x^5& \hfill 0x^4& \hfill 0x^3& \hfill +x^2& \hfill & \hfill \\ \hfill & \hfill & \hfill 2x^4& \hfill -6x^3& \hfill +0x^2& \hfill -5x& \hfill +1\\ \hfill & \hfill & \hfill 2x^4& \hfill +0x^3& \hfill +0x^2& \hfill +2x& \hfill \\ \hfill & \hfill & \hfill & \hfill -6x^3& \hfill +0x^2& \hfill -7x& \hfill +1\\ \hfill & \hfill & \hfill & \hfill -6x^3& \hfill 0x^2& \hfill 0x& \hfill -6\\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill -7x& \hfill +7\end{array}$

The quotient is $x^2+2x-6$ and the remainder is $-7x+7$.

Part 2: No Calculator Allowed

Let’s use synthetic division.

$\begin{array}{rrrrr}\hfill 2& \hfill 1& \hfill -2& \hfill 8& \hfill -4\\ \hfill & \hfill & \hfill 2& \hfill 0& \hfill 16\\ \hfill & \hfill 1& \hfill 0& \hfill 8& \hfill 12\end{array}$

Thus, $g\left(2\right)=12$.

These will be factors of 6 over factors of 2. $\pm \left\{1,2,3,6,\frac{1}{2},\frac{3}{2}\right\}$.

Don’t forget that complex non-real zeros must come in conjugate pairs. $\left(x-3\right)\left(x-\left(2-i\right)\right)\left(x-\left(2+i\right)\right){\left(x+1\right)}^3$

$x^2-4\Rightarrow x=\pm 2$; ${\left(x-4\right)}^2\Rightarrow x=4$ with multiplicity 2; $x^2+1\Rightarrow x=\pm i$.

The possible rational roots are $\pm \left\{1,2,5,10\right\}$. I notice that $p\left(1\right)=1-3-1+13-10=0$, so 1 is a root.

$\begin{array}{rrrrrr}\hfill 1& \hfill 1& \hfill -3& \hfill -1& \hfill 13& \hfill -10\\ \hfill & \hfill & \hfill 1& \hfill -2& \hfill -3& \hfill 10\\ \hfill & \hfill 1& \hfill -2& \hfill -3& \hfill 10& \hfill 0\end{array}$

That makes the factorization $\left(x-1\right)\left(x^3-2x^2-3x+10\right)$. Clearly, neither 1 nor negative 1 will work for the remaining piece…how about 2? $p\left(2\right)=2^3-2\cdot 2^2-3\cdot 2+10=8-8-6+10\ne 0$. Hmmm…maybe negative 2? $p\left(-2\right)={\left(-2\right)}^3-2{\left(-2\right)}^2-3\left(-2\right)+10=-8-8+6+10=0$, so $-2$ is a root.

$\begin{array}{rrrrr}\hfill -2& \hfill 1& \hfill -2& \hfill -3& \hfill 10\\ \hfill & \hfill & \hfill -2& \hfill 8& \hfill -10\\ \hfill & \hfill 1& \hfill -4& \hfill 5& \hfill 0\end{array}$

This makes the factorization $\left(x-1\right)\left(x+2\right)\left(x^2-4x+5\right)$. Time for the quadratic formula!

$x=\frac{4\pm \sqrt{{\left(-4\right)}^2-4\left(1\right)\left(5\right)}}{2\left(1\right)}=\frac{4\pm \sqrt{16-20}}{2}=\frac{4\pm \sqrt{-4}}{2}=\frac{4\pm 2i}{2}=2\pm i$

Thus (finally) we have the list of complex zeros: $1,-2,2\pm i$.

Since the degree is odd, the ends will point in opposite directions. Since the leading coefficient is negative, the right side will point down. Thus, the end behavior is $\left(+\infty ,-\infty \right)$.

Page last updated 14:23 2021-02-16