Mathematics, 29.07.2019 10:30 jewelia2801
Use the complex zeros(1/2, -3+7i, -3-7i) to factor f(x)=-2x^3 - 11x^2 -110x +58
Answers: 2
![\bf f(x)=(1-2x)(x+3+7i)(x+3-7i) \\\\\\ f(x)=(1-2x)[(x+3)+(7i)][(x+3)-(7i)] \\\\\\ f(x)=(1-2x)~~[\stackrel{\textit{difference of squares}}{(x+3)^2-(7i)^2}] \\\\\\ f(x)=(1-2x)[(x^2+6x+9)-(7^2i^2)] \\\\\\ f(x)=(1-2x)[(x^2+6x+9)-(49\cdot -1)] \\\\\\ f(x)=(1-2x)[(x^2+6x+9)-(-49)] \\\\\\ f(x)=(1-2x)[(x^2+6x+9)+49] \\\\\\ f(x)=(1-2x)(x^2+6x+58)](/tpl/images/0146/3585/57bc4.png)
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Use the complex zeros(1/2, -3+7i, -3-7i) to factor f(x)=-2x^3 - 11x^2 -110x +58...
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