Whqt is the equation to x 2+ 12 x - 64=0

Do you want to solve the equation?

Solving by factoring,

x^2 + 12x - 64 = (x+16)(x-4) = 0

x = -16 and 4 <==Answer

Whqt is the equation to x 2+ 12 x - 64=0

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Do you want to solve the equation?

Solving by factoring,

x^2 + 12x - 64 = (x+16)(x-4) = 0

x = -16 and 4 <==Answer

Robert did a shortcut to solve this equation. I will try my best to thoroughly explain it.

x^2 + 12x - 64 = 0 <== The original equation

Some people have trouble factoring and there is a trick to it.

x^2 + 12x **- 64** = 0 <== If you noticed that I bold the number **
64 **it is because I know that there are various of numbers that can result to 64. Such as 2, 4, 8, 16, and 32.

x^2 +** 12x** - 64 = 0<==But I also look at **12**, I know that it have a multiples of 2, 3, 4, and 6.

I will short come back to the equation I want to be aware of ax^2+bx+c=0.

**bx= **know that the multiples of **cx** that you factor out will represent the sum or difference of
**bx**

**cx**= know that they are multiples of some numbers and possibly imaginary (or complex) numbers.

x^2 + 12x - 64 = 0

so we will find the multiples -64 and sum of 12

**2 +/- 32= 34 or 30**

**4 +/- 16= 20 or 12**

**8 +/- 8= 16 or 0**

**
for my cx **= I used the multiples of 64, which are 2, 4, 8, 16, 32 and created a rainbow. By connecting 2 and 32 which is 64, 4 and 16 which is 64, and 8 stays alone because it is multiplied to itself which is 64. As illustrated below.

* for my bx* = I combined them to use them as a sum and difference as illustrated above.

2*32= 64

4*16= 64

8*8= 64

x^2 + 12x - 64 = 0 <== Remember I want to factor them do not get lost

Look above and find the sum or difference of 12 as instructed clearly in this problem. (4 and 16)

the problem tells you there are 2 x's. So I simply obey and separate it.

So now I have ===> (x _ 4)(x _ 16)

x^2 + 12x - 64 = 0==> As you can see we have a slight problem, we need to know if we should make both equations negative, positive, or mixture of both.

However, because I know that 64(cx) is a negative, one of the individuals are too. I look back at my** bx **table and noticed that I found the difference. If 16-4=12 and my 4 is negative and 16 is positive in that simple equation, then
**(x-4)(x+16)** would be my factors. Because I know that 4-16= -12 and that does not match my bx in the original equation.

Now we have to find the zeroes...

x-4=0 x+16=0 <== I separate my factors into 2 different equations.

+4 +4 -16 -16 <== I moved each numbers so I can have x alone

x=4 and x=-16 <=== This is my answer

Hopefully I explained it clear enough for you...