How do you solve #3x ^ { 3} + 18= x ( 2x + 27)#?
2 Answers
Explanation:
First multiply out the bracket on the LHS of the equation.
Rearrange to get a cubic equation.
Now find a number that can be substituted into
When
We now have a factor and now we can factorise the equation.
Now you can find the answers for
Explanation:
I would start by distributing the x on the right side of the equal sign.
Get everything to one side, I moved everything to the left side. Do this by subtracting the terms on the left side. You get
Then reorder into standard form (decrease exponents)
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.
Look for what can be factored out of the first grouping:
Look for what can be factored out of the second groping:
The first grouping now looks like
The second grouping now looks like
Notice the terms in the parenthesis are the same.
your problem now looks like:
Now when you factor by grouping you put the two terms in front of the parenthesis together in front of the set of parenthesis.
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
so x could be +3 or -3.
Don't forget to solve the second equation
Add two to both sides, then divide by 3.