color(blue)("PART 1"PART 1
f(x)=4x^3+16x^2-3x-45f(x)=4x3+16x2−3x−45 Find f(-3)f(−3)
Lets substitute all of xx with color(red)(-3)−3
f(color(red)(-3))=4(color(red)(-3))^3+16(color(red)(-3))^2-3(color(red)(-3))-45f(−3)=4(−3)3+16(−3)2−3(−3)−45
=f(color(red)(-3))=4*(-27)+16*9-(-9)-45=f(−3)=4⋅(−27)+16⋅9−(−9)−45
=f(color(red)(-3))=-108+144+9-45=f(−3)=−108+144+9−45
=f(color(red)(-3))=0=f(−3)=0
color(blue)("PART 2"PART 2
f(x)=4x^3+16x^2-3x-45f(x)=4x3+16x2−3x−45 Find f(x)=0f(x)=0
f(x)=4x^3+16x^2-3x-45=0f(x)=4x3+16x2−3x−45=0
Factorize color(red)(4x^3+16x^2-3x-45->4x3+16x2−3x−45→
f(x)=(x+3)(2x−3)(2x+5)=0f(x)=(x+3)(2x−3)(2x+5)=0
Set the factors equal to zero
color(green)(x+3=0x+3=0 or color(violet)(2x−3=02x−3=0 or color(orange)(2x+5=02x+5=0
therefore color(green)(x=−3 or color(violet)(x=3/2 or color(orange)(x=−5/2