Question

Given f(x)=2x^3−14x^2+bx−3 has zero x=2, and g(x)=x^3+cx^2−8x+30, with c a real number, has zero x=3−i, find (f•g)(−1)

Answer 1

The given functions are

`\begin{gathered} f(x)=2x^3-14x^2+bx-3 \\ g(x)=x^3+cx^2-8x+30 \end{gathered}`

Since f(x) has a zero x = 2, then

Substitute x by 2 and equate the answer by 0 to find b

`\begin{gathered} f(2)=0 \\ 2(2)^3-14(2)^2+b(2)-3=0 \\ 2(8)-14(4)+2b-3=0 \\ 16-56+2b-3=0 \\ (16-56-3)+2b=0 \\ -43+2b=0 \end{gathered}`

Add 43 to both sides

`\begin{gathered} -43+43+2b=0+43 \\ 2b=43 \end{gathered}`

Divide both sides by 2

`\begin{gathered} (2b)/(2)=(43)/(2) \\ b=(43)/(2) \end{gathered}`

Then, f(x) is

`f(x)=2x^3-14x^2+(43)/(2)x-3`

We will do the same with g(x)

Since g(x) has a root x = 3 - i, then there is another root x = 3 + i

Then substitute x by 3 - i and equate the answer by 0

`\begin{gathered} g(x-i)=0 \\ (3-i)^3+c(3-i)^2-8(3-i)+30=0 \\ (3-i)(3-i)^2+c(9-6i+i^2)-24+8i+30=0 \\ (3-i)(9-6i+i^2)+c(8-6i)-24+8i+30=0 \\ 27-18i+3i^2-9i+6i^2-i^3+c(8-6i)-24+8i+30=0 \\ 27-18i-3-9i-6+i+c(8-6i)-24+8i+30=0 \\ (27-3-6-24+30)+(-18i-9i+i+8i)+c(8-6i)=0 \\ 24+(-18i)+c(8-6i)=0 \\ 24-18i+c(8-6i)=0 \\ 8c-6ic=-24+18i \end{gathered}`

By comparing the 2 sides, then

`\begin{gathered} 8c=-24 \\ (8c)/(8)=(-24)/(8) \\ c=-3 \end{gathered}`

Then g(x) is

`g(x)=x^3-3x^2-8x+30`

Now, multiply f and g, then substitute x by -1

`(f\ast g)(-1)=\lbrack2(-1)^3-14(-1)^2+(43)/(2)(-1)-3\rbrack\ast\lbrack(-1)^3-3(-1)^2-8(-1)+30\rbrack`

Simplify each bracket

`\begin{gathered} (f\ast g)(-1)=\lbrack-2-14-(43)/(2)-3\rbrack\lbrack-1-3+8+30\rbrack \\ (f\ast g)(-1)=\lbrack-(77)/(2)\rbrack\ast\lbrack34\rbrack \\ (f\ast g)(-1)=-1309 \end{gathered}`

The answer is -1309