Trig - factoring

[dad-4]

I am trying to help my son with his trig homework and I am completely lost. He is trying to factor. An example of one of his problems is:

f(x) = 5x^3 - 9x^2 + 28x + 6

The book gives us the answer -1/5, 1, +/-1/2i

I am trying to help him with how to solve the problem. We understand basic factoring e.g., x^2 + x - 12 = (x+4) * (x-3) , but I do not know how to approach the problem above.

 

[arthur ohlsten]

There is something wrong with the problem
y=5x^3 - 9x^2+28x+6

this is a third order equation, and only has 3 roots,not 4.

Also at x=1 y=30 not 0 as it would if 1 was a root

roots you give are- 1/5, 1, +/- i/2 if these are the roots then the equation would be:
y=[x+1/5][x-1][x+i/2][x-i/2] expanding
y=[x^2-4/5x -1/5][x^2+1/4]
y={1/[5*4] }[5x^2-4x-1][4x^2+1]
y=[1/20] [ 20x^4-16x^3+x^2-4x-1]
y=x^4-4x^3/5 +x^2/20 -x/5-1/20

please check the problem in the text.
Arthur

 

[soroban]

Hello, dad-4!

Arthur is absolutely correct.
Those answers do not belong to that equation.

\(\displaystyle f(x) \:= \:5x^3 \,-\, 9x^2\, +\, 28x\, +\, 6\)

I assume that the problem says something like:
. . Find the zeros of: \(\displaystyle \:f(x)\:=\:5x^3\,-\,9x^2\,+\,28x\,+\,6\:=\:0\)

If you're not familiar with the Factor Theorem and Remainder Theorem,
. . this is an awful problem.
[Perhaps someone will explain them to you.]

After a while, we find that \(\displaystyle x\,=\,-\frac{1}{5}\) is a zero of \(\displaystyle f(x).\)

This means: \(\displaystyle \,\left(x\,+\,\frac{1}{5}\right) \:\Rightarrow\5x\,+\,1)\) is a factor.

Using long (or synthetic) division: \(\displaystyle \:f(x) \:=\5x\,+\,1)(x^2\,-\,2x\,+\,6)\)

Use the Quadratic Formula on the second factor: \(\displaystyle \:x\:=\:\frac{2\,\pm\,\sqrt{(-2)^2\,-\,4(1)(6)}}{2(1)} \:=\:\frac{2\,\pm\,\sqrt{-20}}{2}\)

. . and we have: \(\displaystyle \:x\:=\:\frac{2\,\pm\,2\sqrt{5}i}{2} \:=\:1\,\pm\,i\sqrt{5}\)


The three zeros are: \(\displaystyle \,-\frac{1}{2},\;1\,\pm\,i\sqrt{5}\)