Factoring
In the following polynomial,
3x(6x² - 17x + 12)
We have to factor the second term of the multiplication in order to have it completely factorized
Factoring 6x² - 17x + 12
We want to factor it by grouping
Step 1
We split the middle term:
since
-17 = -9 -8, then
6x² - 17x + 12 = 6x² - 8x - 9x + 12
Step 2
We group into pairs:
6x² - 8x - 9x + 12 = (6x² - 8x) - (9x - 12)
Step 3
We factor each bynomial:
First bynomial 6x² - 8x:
Since
2 · 3 = 6
2 · 4 = 8
then
6x² - 8x = 2 · 3x² - 2 · 4x
We can observe that 2 · 3x² and 2 · 4x have two common factors: 2 and x. Then
6x² - 8x = 2x(3x - 4)
Second bynomial 9x - 12:
Since
3 · 3 = 9
3 · 4 = 12
then
9x - 12 = 3 · 3x - 3 · 4
We can observe that 3 · 3x and 3 · 4 have one common factor: 3. Then
9x - 12 = 3(3x - 4)
Step 4
We can replace the results found in the last step:
6x² - 17x + 12
= (6x² - 8x) - (9x - 12)
= 2x(3x - 4) - 3(3x - 4)
We observe that 2x(3x - 4) and 3(3x - 4) have a common factor: (3x - 4)
Then, we can factorise:
= 2x(3x - 4) - 3(3x - 4)
= (3x - 4)(2x - 3)
Therefore
6x² - 17x + 12 = (3x - 4)(2x - 3)
Using this we can replace in the original expression
Answer: 3x(6x² - 17x + 12) = 3x(3x - 4)(2x - 3)