Question

pre - calculus ii
2.2 polynomials of a higher degree
match the possible equation with the given graph:

1. $f(x)=-2x + 3$
2. $f(x)=-\frac{1}{4}x^{4}+3x^{2}$
3. $f(x)=-2x^{2}-5x$
4. $f(x)=\frac{1}{5}x^{5}-2x^{3}+\frac{9}{5}x$
5. $f(x)=x^{2}-4x$
6. $f(x)=2x^{3}-3x + 1$
7. $f(x)=-\frac{1}{3}x^{3}+x^{2}-1$
8. $f(x)=x^{4}+2x^{3}$

#1 - 17 no graphing calculator
for each problem, find all the real zeros of the polynomial functions by factoring. determine the multiplicity of each zero.

1. $f(x)=x^{4}-x^{3}-20x^{2}$
2. $f(x)=x^{4}-10x^{2}+9$
3. $f(x)=x^{3}+2x^{2}-16x - 32$
4. $f(x)=2x^{4}-14x^{2}+24$

Explanation:

Step1: Factor the polynomial $f(x)=x^{4}-x^{3}-20x^{2}$

Factor out the greatest - common factor $x^{2}$ first: $f(x)=x^{2}(x^{2}-x - 20)$. Then factor the quadratic $x^{2}-x - 20=(x - 5)(x+4)$. So $f(x)=x^{2}(x - 5)(x + 4)$.

Step2: Find the zeros

Set $f(x)=0$. Then $x^{2}(x - 5)(x + 4)=0$. Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $x^{2}=0$ gives $x = 0$ with multiplicity 2, $x-5=0$ gives $x = 5$ with multiplicity 1, and $x + 4=0$ gives $x=-4$ with multiplicity 1.

Step3: Factor the polynomial $f(x)=x^{4}-10x^{2}+9$

Let $u=x^{2}$, then the polynomial becomes $u^{2}-10u + 9$. Factor $u^{2}-10u + 9=(u - 1)(u - 9)$. Substitute back $u=x^{2}$, we get $f(x)=(x^{2}-1)(x^{2}-9)=(x - 1)(x + 1)(x - 3)(x + 3)$.

Step4: Find the zeros

Set $f(x)=0$. Using the zero - product property, $x-1=0$ gives $x = 1$ with multiplicity 1, $x + 1=0$ gives $x=-1$ with multiplicity 1, $x - 3=0$ gives $x = 3$ with multiplicity 1, and $x + 3=0$ gives $x=-3$ with multiplicity 1.

Step5: Factor the polynomial $f(x)=x^{3}+2x^{2}-16x - 32$

Group the terms: $(x^{3}+2x^{2})-(16x + 32)=x^{2}(x + 2)-16(x + 2)=(x + 2)(x^{2}-16)=(x + 2)(x - 4)(x + 4)$.

Step6: Find the zeros

Set $f(x)=0$. Using the zero - product property, $x + 2=0$ gives $x=-2$ with multiplicity 1, $x - 4=0$ gives $x = 4$ with multiplicity 1, and $x + 4=0$ gives $x=-4$ with multiplicity 1.

Step7: Factor the polynomial $f(x)=2x^{4}-14x^{2}+24$

Factor out the greatest - common factor 2 first: $f(x)=2(x^{4}-7x^{2}+12)$. Let $u=x^{2}$, then $u^{2}-7u + 12=(u - 3)(u - 4)$. Substitute back $u=x^{2}$, we get $f(x)=2(x^{2}-3)(x^{2}-4)=2(x-\sqrt{3})(x+\sqrt{3})(x - 2)(x + 2)$.

Step8: Find the zeros

Set $f(x)=0$. Using the zero - product property, $x-\sqrt{3}=0$ gives $x=\sqrt{3}$ with multiplicity 1, $x+\sqrt{3}=0$ gives $x=-\sqrt{3}$ with multiplicity 1, $x - 2=0$ gives $x = 2$ with multiplicity 1, and $x + 2=0$ gives $x=-2$ with multiplicity 1.

Answer:

For $f(x)=x^{4}-x^{3}-20x^{2}$, the zeros are $x = 0$ (multiplicity 2), $x = 5$ (multiplicity 1), $x=-4$ (multiplicity 1).
For $f(x)=x^{4}-10x^{2}+9$, the zeros are $x = 1$ (multiplicity 1), $x=-1$ (multiplicity 1), $x = 3$ (multiplicity 1), $x=-3$ (multiplicity 1).
For $f(x)=x^{3}+2x^{2}-16x - 32$, the zeros are $x=-2$ (multiplicity 1), $x = 4$ (multiplicity 1), $x=-4$ (multiplicity 1).
For $f(x)=2x^{4}-14x^{2}+24$, the zeros are $x=\sqrt{3}$ (multiplicity 1), $x=-\sqrt{3}$ (multiplicity 1), $x = 2$ (multiplicity 1), $x=-2$ (multiplicity 1).