Factor completely. Remember to look first for a common factor and check by multiplying. if polynomial is prime, state this.
-36a^2-96ab-54b^2
I couldn't find the right answer.
HCF of -36, -96, and -54 is -6
I see no same variable in all 3 terms
-36a^2-96ab-54b^2
= -6(6a^2 + 16ab + 9b^2)
the trinomial does not factor further over the rationals.
Mentally expand to check the answer.
Hello Reiny,
I am sorry I have a typo. The problem should be -36a^2-96ab-64b^2.
I am unsure of the steps.
Again first look for a common factor in the constants which is -4
-36a^2-96ab-64b^2
= -4(9a^2 + 24ab + 16b^2)
the 9a^2 and 16b^2 are both perfect squares, could we perhaps have (3a + 4b)^2 ?
A quick mental expansion shows this is indeed the case, so
-36a^2-96ab-64b^2
= -4(9a^2 + 24ab + 16b^2)
= -4(3a + 4b)^2
To factor the given polynomial, -36a^2 - 96ab - 54b^2, we need to look for common factors, such as -6, which can be factored out. By doing this, we divide each term in the polynomial by -6:
-6(6a^2 + 16ab + 9b^2)
Now, let's see if we can further factor the expression inside the bracket, 6a^2 + 16ab + 9b^2. This is a trinomial, so we need to check if it can be factored. Let's consider the factors of the quadratic term (6a^2) and the constant term (9b^2), which are 6a and 3b respectively.
By constructing a pair of parentheses, we can rearrange the expression using these factors:
(2a + 3b) (3a + 3b)
Notice that the two terms inside the parentheses, (2a + 3b) and (3a + 3b), have a common factor of (3b). Factoring out this common factor, we get:
3b(2a + 3b + 3a + 3b)
Now, we can simplify the expression further:
3b(5a + 6b)
Therefore, the fully factored form of the given polynomial -36a^2 - 96ab - 54b^2 is -6 * 3b(5a + 6b).
Please note that if factoring is not possible, then the polynomial is considered prime. In this case, we have successfully factored the polynomial.