Question

Factor the polynomial f(x). Then solve the equation f) o. 10) f(x) x3+5x2- 9x-45 State the domain of the rational function. (6 points) 11) g(x) =ー2 x +2 Given that the polynomial function has the given zero, find the other zeros. 12) f(x)=x3-4x2 + 9x-10:2

Answer 1

Solution : ( 10 )

$f(x)=x^3+5x^2-9x-45$

$\mathrm{Factor}\:x^3+5x^2-9x-45:$

$\mathrm{Steps\::}$

$x^3+5x^2-9x-45$

$=\left(x^3+5x^2\right)+\left(-9x-45\right)$

$\mathrm{Factor\:out\:}x^2\mathrm{\:from\:}x^3+5x^2\mathrm{:\quad }x^2\left(x+5\right)$

$\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9x-45\mathrm{:\quad }-9\left(x+5\right)$

$=x^2\left(x+5\right)-9\left(x+5\right)$

$\mathrm{Factor\:out\:common\:term\:}\left(x+5\right)$

$=\left(x+5\right)\left(x^2-9\right)$

$=\left(x+5\right)\left(x^2-3^2\right)$

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$=\left(x+5\right)\left(\left(x+3\right)\left(x-3\right)\right)$

$=\left(x+5\right)\left(x+3\right)\left(x-3\right)$

$\mathrm{We\:have\:now\:factored\:the\:polynomial\:completely.}$

$f(x)=\left(x+5\right)\left(x+3\right)\left(x-3\right)$

$\mathrm{Now\:that\:we\:have\:factored\:}f\left(x\right),\mathrm{\:the\:Zero\:Product\:Property\:tells\:us\:that\:the\:}$

$\mathrm{solutions\:to\:}f(x)=0\:\mathrm{\:are\:the\:values\:of\:}x\mathrm{\:that\:make\:a\:factor\:zero.\:So:}$

$x+5=0\mathrm{\:\:or\:\:}x+3=0\mathrm{\:\:or\:\:}x-3=0$

$\mathrm{Solving\:each\:of\:these\:for\:}x\mathrm{\:we\:get\::}$

$x=-5\mathrm{\:\:or\:\:}x=-3\mathrm{\:\:or\:\:}x=3$

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Solution : ( 11 )

$g\left(x\right)=\frac{x-2}{x+2}$

$\mathrm{Domain\:of\:}\:\frac{x-2}{x+2}\::\quad$

$\mathrm{Steps\::}$

$\mathrm{Find\:undefined\:\left(singularity\right)\:points}:\quad$

$\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{x-2}{x+2}\mathrm{\:and\:compare\:to\:zero}$

$\mathrm{Solve\:}\:x+2=0:\quad$

$x+2=0$

$\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}$

$x+2-2=0-2$

$\mathrm{Simplify}$

$x=-2$

$\mathrm{The\:following\:points\:are\:undefined}$

$x=-2$

$\mathrm{The\:function\:domain}$

$x<-2\quad \mathrm{or}\quad \:x>-2$

$\mathrm{Hence,}$

$\mathrm{Domain\:of\:}\:\frac{x-2}{x+2}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x<-2\quad \mathrm{or}\quad \:x>-2\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:-2\right)\cup \left(-2,\:\infty \:\right)\end{bmatrix}$

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Solution : ( 12 )

$f(x)=x^3-4x^2+9x-10$

$\mathrm{Steps \::}$

$x^3-4x^2+9x-10=0$

$\mathrm{Solve\:by\:factoring}$

$\mathrm{Factor\:}x^3-4x^2+9x-10:$

$x^3-4x^2+9x-10$

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=10,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:5,\:10,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:5,\:10}{1}$

$\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-2$

$\mathrm{Compute\:}\frac{x^3-4x^2+9x-10}{x-2}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }x^2-2x+5$

(z-2) (2.2-22, +5)

(z-2) (2.2-2x + 5) = 0

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

From (1);

$\mathrm{Solve\:}\:x-2=0:\quad x=2$

$\mathrm{and}$

$\mathrm{Similarly.\:\:From\:\:(1)};$

$\mathrm{Solve\:}\:x^2-2x+5=0:\quad$

$x^2-2x+5=0$

Solve with the quadratic formula

For a quadratic equation of the form ax2 + bx + c = 0 the solutions are

ас C0

$\mathrm{For\:}\quad a=1,\:b=-2,\:c=5:\quad x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4*1*5}}{2*1}$

$x_{1}=\frac{-\left(-2\right)+ \sqrt{\left(-2\right)^2-4*1*5}}{2*1}$

$=\frac{2+\sqrt{16}i}{2}$

$=\frac{2+4i}{2}$

$=\frac{2\left(1+2i\right)}{2}$

$=1+2i$

$\mathrm{and}$

$x_{2}=\frac{-\left(-2\right)- \sqrt{\left(-2\right)^2-4*1*5}}{2*1}$

$=\frac{2-\sqrt{16}i}{2}$

$=\frac{2-4i}{2}$

$=\frac{2\left(1-2i\right)}{2}$

$=1-2i$

The final solutions to the quadratic equation are

$x=1+2i,\:x=1-2i$

The final solutions to the equation are:

$x=2,\:x=1+2i,\:x=1-2i$

$\mathrm{\:Hence,\:\:the\:other\:two\:zeros}$

$x=1+2i,\:x=1-2i$