PreCalculus Honors Unit Test #2

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Part 1: Calculator Allowed

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

 1 Locate intervals where $f\left(x\right)$ is increasing and decreasing. (5)

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

 2 Locate and classify the extrema of $f\left(x\right)$. (4)

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

 3 Locate all real zeros of $f\left(x\right)$. (4)

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

 4 Find the quotient and remainder: $\frac{{x}^{5}+2{x}^{4}-6{x}^{3}+{x}^{2}-5x+1}{{x}^{3}+1}$ (5)

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

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

 PreCalculus Honors Unit Test #2

Part 2: No Calculator Allowed

 5 Use the remainder theorem to find $g\left(2\right)$ if $g\left(x\right)={x}^{3}-2{x}^{2}+8x-4$. (4)

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

 6 List the possible rational roots of $f\left(x\right)=2{x}^{5}-{x}^{4}-10{x}^{3}+5{x}^{2}+12x-6$. (4)

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

 7 Write a polynomial that has the following zeros and multiplicities: 3, $2-i$, $-1$ (multiplicity 3) (5)

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

 8 List all complex zeros (and multiplicities) of $h\left(x\right)=\left({x}^{2}-4\right){\left(x-4\right)}^{2}\left({x}^{2}+1\right)$. (6)

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

 9 Find all complex zeros of $p\left(x\right)={x}^{4}-3{x}^{3}-{x}^{2}+13x-10$. (7)

The possible rational roots are $±\left\{1,2,5,10\right\}$. 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 $\left(x-1\right)\left({x}^{3}-2{x}^{2}-3x+10\right)$. 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!

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

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

 10 Determine the end behavior of $q\left(x\right)=-2{x}^{5}-{x}^{4}+3{x}^{3}-{x}^{2}-x+1$. (3)

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 $\left(+\infty ,-\infty \right)$.

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