How do you solve 3x3+18=x(2x+27)?

2 Answers
Jun 15, 2017

x=3,3,23

Explanation:

First multiply out the bracket on the LHS of the equation.

3x3+18=2x2+27x

Rearrange to get a cubic equation.

3x3+182x227x=0

3x32x227x+18=0

Now find a number that can be substituted into x that will make the equation equal zero.

When x = 3, the equation is zero.

We now have a factor and now we can factorise the equation.

(x3)(3x2+7x6)(x3)(3x2)(x+3)

Now you can find the answers for x.

x=3,3,23

Jun 15, 2017

x=3,3,23

Explanation:

I would start by distributing the x on the right side of the equal sign.

3x3+18=2x2+27x

Get everything to one side, I moved everything to the left side. Do this by subtracting the terms on the left side. You get

3x3+182x227x=0

Then reorder into standard form (decrease exponents)

3x32x227x+18=0

When ever you have 4 terms, it is usually a factoring by grouping problem. Put parenthesis around the first two and the last two terms.
(3x32x2)(27x+18)=0

Look for what can be factored out of the first grouping: x2
Look for what can be factored out of the second groping: 9

The first grouping now looks like x2(3x2)
The second grouping now looks like 9(3x2)

Notice the terms in the parenthesis are the same.

your problem now looks like:
x2(3x2)9(3x2)=0

Now when you factor by grouping you put the two terms in front of the parenthesis together in front of the set of parenthesis.

(x29)(3x2)=0

Now set each sit of parenthesis equal to zero and solve

#(x^2-9) = 0 and (3x-2) =0.

I recognized the first terms as the difference of two squares which means it can be easily factor
(x29)(x+3)(x3)=0

so x could be +3 or -3.

Don't forget to solve the second equation 3x2=0
Add two to both sides, then divide by 3.

x=23

x=3,3,23