x^2 - 8x + 16 = 0
(x )(x )
You need two numbers that when multiplied are equal to 16, and when added are equal to 8.
These two numbers are 4 and 4 (4 * 4 = 16, 4 + 4 = 16)
Then you need to figure out if these are positive or negative. Since 16 is positive, you know that the numbers are either both positive or both negative. Since the 8 is negative, you know they must be negative.
(x - 4)*(x - 4)
Find factors of 16 which add to 8: Factors of 16 are 1 & 16, 2 & 7 and 4 & 4: Only the last add to 8 so: (x + 4)(x + 4) or (x + 4)2
That factors to (x^3 + 64)(x^3 + 64) which factors to (x + 4)(x + 4)(x^2 - 4x + 16)(x^2 - 4x + 16)
(x + 8)(x + 2)
It is 112
It equals 113.
The factorization of 25x2 + 40x + 16 is (5x+4)(5x+4).
-((x + 2)(x - 8))
(2xy-4x)+ (8y-16)=2xy-4x+8y-16, which factors as 2(x+4)(y-2)
Find factors of 16 which add to 8: Factors of 16 are 1 & 16, 2 & 7 and 4 & 4: Only the last add to 8 so: (x + 4)(x + 4) or (x + 4)2
That factors to (x^3 + 64)(x^3 + 64) which factors to (x + 4)(x + 4)(x^2 - 4x + 16)(x^2 - 4x + 16)
(x + 8)(x + 2)
All numbers have factors. Some numbers have some of the same factors as other numbers. Multiples of numbers have all of the factors of numbers that are factors of them, plus some of their own. 32 is a multiple of 16 and will have all of its factors. 64 is a multiple of both 32 and 16 and will have all of their factors.
It is 112
96
It equals 113.
The answer is 192. Simply add up all of the numbers arithmetically like you would for two plus two. To make sure the answer is correct, check your work using a calculator.
x3 + 4x2 + 16x + 64 =x2*(x + 4) + 16(x + 4) = (x + 4)*(x2 + 16) which has no further real factors.