Determine the equation of the following polyromial that passes through point (0,6).

Cubic polynomial has zeros at x=−1x=−1 and 22, is tangent to x−x−axis at x=−1x=−1, and passes through the point (0,−6)(0,−6).

So cubic polynomial has double zero at x=−1x=−1, and single zero at x=2x=2

f(x)=a(x+1)2(x−2)f(x)=a(x+1)2(x−2)

f(0)=−6f(0)=−6

a(1)(−2)=−6a(1)(−2)=−6

a=3a=3

f(x)=3(x+1)2(x−2)f(x)=3(x+1)2(x−2)

f(x)=3x3−9x−6

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4

5

step-by-step explanation: