Which expression correctly shows p(x)=9x5−9x2 factored completely over the integers?
9x2(x2+1)(x2−1)
(x−1)(x2+x+1)
9x2(x−1)(x2+x+1)
9x2(x2−2x+1)
is it (x−1)(x2+x+1)
Did you check your answer? It's not even a 5th degree polynomial.
You should have stopped here:
9x^2(x−1)(x^2+x+1)
The next line is bogus, since x^2-2x+1 = (x-1)^2
And you can't just discard factors as you go.
Oh okay I understand what you are saying thank you. I am really struggling in this subject no matter what videos I watch and go over my notes I just cant get it
Multiply. (3x + 5)(3x – 5)
A. 9x2 – 30x – 25
B. 9x2 + 30x + 25
C. 9x2 – 25
D. 9x2 + 25
The given expression is a difference of squares and can be written as:
(3x + 5)(3x - 5) = (3x)^2 - 5^2
= 9x^2 - 25
Therefore, the answer is (C) 9x2 – 25.
To determine which expression correctly shows \(p(x)=9x^5-9x^2\) factored completely over the integers, we need to find the factors of the polynomial.
One approach is to factor out the greatest common factor first. In this case, the greatest common factor is \(9x^2\). Factoring it out, we get:
\[p(x) = 9x^2(x^3-1)\]
Next, we can notice that \(x^3-1\) can be factored as a difference of cubes, because it follows the formula \(a^3-b^3 = (a-b)(a^2+ab+b^2)\) where \(a = x\) and \(b = 1\). Applying this formula, we have:
\[p(x) = 9x^2(x-1)(x^2+x+1)\]
Therefore, the correct expression that shows \(p(x)=9x^5-9x^2\) factored completely over the integers is:
\[9x^2(x-1)(x^2+x+1)\]
Thus, the expression (x-1)(x^2+x+1) is indeed part of the correct factored form, but it is missing the greatest common factor of 9x^2.