Which expression correctly shows p(x)=27x3+45x2−3x−5 factored completely over the integers? (Hint: You will need to use more than one method to complete this problem.)
(3x−5)(9x2−3x+9)
(3x−5)(9x2+3x+9)
(3x+5)(3x+1)(3x−1)
(3x+5)(9x2−1)
almost.
9x^2-1 = (3x)^2 - 1^2 = (3x-1)(3x+1)
oh okay so I converted the two 3x to far. Thank you for your help!
To factor the expression p(x) = 27x^3 + 45x^2 - 3x - 5 completely over the integers, we can use a combination of methods.
Method 1: Factor out the greatest common factor (GCF)
The GCF of the expression is 1 since there are no common factors other than 1. So we do not need to factor out a GCF in this case.
Method 2: Factor by grouping
For the expression p(x) = 27x^3 + 45x^2 - 3x - 5, we can group the terms as follows:
Group 1: (27x^3 + 45x^2)
Group 2: (-3x - 5)
In Group 1, we can factor out the GCF of 9x^2:
Group 1: 9x^2(3x + 5)
In Group 2, we can factor out the GCF of -1:
Group 2: -1(3x + 5)
Now we have:
p(x) = 9x^2(3x + 5) - 1(3x + 5)
Notice that we now have a common factor of (3x + 5) in both terms. We can factor it out:
p(x) = (3x + 5)(9x^2 - 1)
Now, let's proceed with factoring the quadratic expression (9x^2 - 1).
Method 3: Recognize the difference of squares pattern
The expression 9x^2 - 1 is in the form of a^2 - b^2, which is a difference of squares. We can use the formula (a + b)(a - b) to factor it:
9x^2 - 1 = (3x)^2 - 1^2 = (3x + 1)(3x - 1)
Therefore, the factored form of p(x) is:
p(x) = (3x + 5)(9x^2 - 1)
Out of the given answer choices, the correct expression that shows the completely factored form of p(x) is:
(3x + 5)(9x^2 - 1)