Factor completely: 3x^2(4x2 + 1)^8 + 64x^4(4x2 + 1)^7.
11 years ago
11 years ago
how did you solve this??? I am so los.
11 years ago
3x^2(4x2 + 1)^8 + 64x^4(4x2 + 1)^7
Notice that x^2 is a factor of each term, and so is a power of (4x^2+1)
So, factoring them out, we have
x^2 (4x^2+1)^7 (3(4x^2+1)+64x^2)
3(4x^2+1)+64x^2 = 12x^2+3+64x^2 = 76x^2+3
and that's Tony's answer
11 months ago
To factor the given expression completely, we can start by factoring out the common factor, which is (4x^2 + 1)^7.
Let's rewrite the expression:
3x^2(4x^2 + 1)^8 + 64x^4(4x^2 + 1)^7
Now we can factor out (4x^2 + 1)^7 as a common factor:
(4x^2 + 1)^7(3x^2(4x^2 + 1) + 64x^4)
Next, let's simplify the expression inside the parentheses:
3x^2(4x^2 + 1) + 64x^4
Multiplying 3x^2 by 4x^2, we get 12x^4, and multiplying 3x^2 by 1, we get 3x^2. Therefore, we have:
12x^4 + 3x^2 + 64x^4
Combining like terms, we get:
76x^4 + 3x^2
Finally, we can rewrite the factored expression as:
(4x^2 + 1)^7(76x^4 + 3x^2)
So, the expression 3x^2(4x^2 + 1)^8 + 64x^4(4x^2 + 1)^7 is factored completely as (4x^2 + 1)^7(76x^4 + 3x^2).