What are the steps to solving both parts correctly?

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Answer 1 · 2017-10-18T05:39:58

$PART 1$ $color(blue)("PART 1"$

$f (x) = 4 x^{3} + 16 x^{2} - 3 x - 45$ $f(x)=4x^3+16x^2-3x-45$ Find $f (- 3)$ $f(-3)$

Lets substitute all of $x$ $x$ with $- 3$ $color(red)(-3)$

$f (- 3) = 4 {(- 3)}^{3} + 16 {(- 3)}^{2} - 3 (- 3) - 45$ $f(color(red)(-3))=4(color(red)(-3))^3+16(color(red)(-3))^2-3(color(red)(-3))-45$

$= f (- 3) = 4 \cdot (- 27) + 16 \cdot 9 - (- 9) - 45$ $=f(color(red)(-3))=4*(-27)+16*9-(-9)-45$

$= f (- 3) = - 108 + 144 + 9 - 45$ $=f(color(red)(-3))=-108+144+9-45$

$= f (- 3) = 0$ $=f(color(red)(-3))=0$

$PART 2$ $color(blue)("PART 2"$

$f (x) = 4 x^{3} + 16 x^{2} - 3 x - 45$ $f(x)=4x^3+16x^2-3x-45$ Find $f (x) = 0$ $f(x)=0$

$f (x) = 4 x^{3} + 16 x^{2} - 3 x - 45 = 0$ $f(x)=4x^3+16x^2-3x-45=0$

Factorize $4 x^{3} + 16 x^{2} - 3 x - 45 \to$ $color(red)(4x^3+16x^2-3x-45->$

$f (x) = (x + 3) (2 x - 3) (2 x + 5) = 0$ $f(x)=(x+3)(2x−3)(2x+5)=0$

Set the factors equal to zero

$x + 3 = 0$ $color(green)(x+3=0$ or $2 x - 3 = 0$ $color(violet)(2x−3=0$ or $2 x + 5 = 0$ $color(orange)(2x+5=0$

$therefore color(green)(x=−3$ or $color(violet)(x=3/2$ or $color(orange)(x=−5/2$