How would I solve this equation using factoring? x^2 + 6 = -5x ?

Question

How would I solve this equation using factoring? and how would i show my steps?

x^2 plus 6 = -5x

Answer 1

It's been a while but I think this is right:

First I put all the numbers on one side. The -5x becomes +5x when it crosses the = sign.

x^2 + 5x + 6 = 0

Then you divide the equasion in two brackets.

(x + __ ) * (x + __ ) = 0

Then you put in the numbers.

To get the number close to x (in this case 5x) you have to sum the numbers (2 + 3 = 5)

To get the number that stays alone (in this case 6) you have to multiply

the numbers ( 2 * 3 = 6 )

(x + 2 ) * (x + 3 ) = 0

If any of the factors is 0 then the eqasion is 0 also.

X = -2

(-2 + 2 ) * ( -2 + 3 ) = 0

0 * 1 = 0

0 = 0

X = -3

( -3 + 2 ) * ( -3 + 3) = 0

-1 * 0 = 0

0 = 0

Answer 2

x^2+6=-5x

x^2+5x+6=0 move the 5x to the other side

(x+3)(x+2) factor

x+3=0  x+2=0 set each one equal to zero

x=-3     x=-2 solve for x

plug them in to see if they work

Answer 3

x^2 + 6 = -5x

set equation to 0

and follow this format

ax^2 + bx + c

x^2 + 5x + 6 = 0

now factor into a binomial

(x + 3) (x + 2)

check ?

use F.O.I.L. (first, outer, inner, last)

first (x)(x) = x^2

outer(2)(x) = 2x

inner (3)(x) = 3x

last (3)(2) = 6

x^2 + 2x + 3x + 6

x^2 + 5x + 6 = 0  

when you have the equation in this form : x^2 + 5x + 6 = 0  

you have to ask yourself what two numbers can you multiply to get positive 6 and you can also use the same ones to add and get +5

these are 3 , and 2 (3x2)= 6 (3+2) = 5

6 is C and 5 is B

ax^2 + bx + c

hope this helps !

Answer 4

x^2 + 6 = -5x

x^2 + 5x + 6 = 0

x^2 + 3x + 2x + 6 = 0

x ( x +3 )+ 2 ( x + 3) =0

( x + 3) ( x + 2 ) = 0

x + 3 = 0 or x + 2 = 0

x = -3 or -2

Answer 5

1:   (x^2)+5x+6=0

2:   (x+3)(x+2)=0

3:   x=-3   OR   x=-2

(using S & P) (S=5, P=6)