Factor completely. 8x^2 - 16x - 64?

Answer 1

First: find the least common multiple or, a number that's divisible with the terms in the expression (8).

8(x^2 - 2x - 8)
x^2 - 2x - 8

Sec: factor the expression > multiply the 1st & 3rd term to get (- 8). find two numbers that give you (- 8) when multiplied &
(- 2) when added/subtracted. the numbers are (- 4 & 2). Rewrite the expression with the new middle terms.

x^2 - 4x + 2x - 8

Third: with 4 terms - group "like" terms & factor both sets of parenthesis.

(x^2 + 2x) - (4x - 8)
x(x + 2) - 4(x + 2)
(x + 2)(x - 4)

Now, combine the terms with the least common multiple (8).

8(x + 2)(x - 4)

Answer 2

In any expression of the type ax^2+bx+c ,for factorisation split the middle term coefficient 'b' into two numbers x,y such that
x*y=a*c;
x+y=b.
Here a=8; b=-16; c=-64
ac=8*-64=-512
So 'b' that is -16 shold be split into 16 and -32 ( you need practice to get this step)
Now, 8x^2-16x-64=8x^2+16x-32x-64
=8x(x+2)-32(x+2)
=(8x-32)(x+2)
=8(x-4)(x+2)
Factors are:
8,x-4.x+2

Answer 3

Careful about those who say you can remove the 8. Since it is not an equation the 8 must remain part of the "expression"

Step one: always factor out anything that is common - in this case the 8 gets factored out but must remain outside the paranthesis. 8(x^2 - 2x - 8)

Step two: Since the a position = 1, the b position is negative and the c position is negative {from ax^2 +bx + x} the factored form will take the form 8(x + )(x - )

Step three: Find two numbers whose product is -8 and whose sum is - 2. For instance the factors for -8 would be:
1, -8 2, -4 -1, 8 -2, 4
but only 2, -4 will be equal to -2 when added, so these are the two numbers which will be put in the empty spaces:

8(x + 2)(x - 4)

Answer 4

8x² - 16x - 64= 0
8(x² - 2x - 8) = 0
8(x - 4)(x +2) = 0
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Answer 5

8(x+4)(x-2)

Answer 6

8x^2-16x-64
=8(x^2-2x-8)
=8(x-4)(x+2)

Answer 7

I know how to do those equations
and I think that is actually impossible - sorry
i tried
but to do those that have a nuber before the x^2 you need to multiply that number in front of the x^2 by the -32
and so on
i'l just confise you and i'm not helping
i'm sorry
But i think one f the signs might be wrong or something because i think that is IMPOSSIBLE!

Answer 8

8(x-4)(x+2)

Answer 9

Reduce by 8 first; leaves you w/
x^2-2x-8
(x-4)(x+2)

Answer 10

obviously we can take out 8 producing 8(x^2-2x-8)
which factorises into 8(x-4)(x+2)