solve the following by factioring x^2-6x-16=0?

x² - 6x - 16 = 0

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Factorisation
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When we have a function in the form of ax² + bx + c, to factorise we use the following steps:

Multiply a by c:

1 × -16 = -16

What we need to do is find what multiplies to make -16, but adds to give the middle term (b [-6] ). To do this we make a list of the factors until we find the right one:

1 × -16
2 × -8 <- This one adds to -6

We have two values: +2 and -8.

Method [ 1 ]:

Substitute these two values for our middle term (-6x), but as factors of x ( 2x and -8x ):

x² + 2x - 8x - 16 = 0

Imagine we have it split into two different parts:

x² + 2x | - 8x - 16 = 0

Factorise each part:

x( x + 2 ) -8( x + 2 ) = 0

*Note: if the two bracketed terms are not the same, you have done something wrong.

Now, we take one of the bracketed terms and the ubracketed terms:

( x + 2 )( x - 8 ) = 0

Method[ 2 ]:

Substitute this into the following unknown brackets:

( x _ ___ )( x _ ___ ) = 0

( x + 2 )( x - 8 ) = 0

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Now we can solve:

x + 2 = 0
x = -2

x - 8 = 0
x = 8

x = -2, 8

Note: Method[ 1 ] is necessary when term a is larger than 1 (i.e. 2x² or -2x² ). When term a is only 1 or -1, Method[ 2 ] will suffice.

x^2-6x-16=0 p=-16
x^2-8x+2x-16=0 s=-6
x(x-8)+2 (x-8)
(x+2)(x-8)
therefore x=-2 or x=8 F=-8, 2

x^2 - 6x - 16 =0
x^2 + 2x - 8x - 16 =0
x(x + 2) -8(x + 2)=0
(x - 8)(x + 2)=0
x - 8=0
x=8

x + 2=0
x= -2

since this is a second order quadratic equation ,
x^2-6x-16= (x-8)(x+2)=0
therefore either x-8=0, or x+2=0
ie.. x=8, x=-2

you must do yourself first

( x - 8 ) ( x + 2 ) = 0

x = 8 , x = - 2