How do you factorise x^2 + 8x + 12?

1:SPLIT THE MID FACTOR
X^2+8X+12 EQUALS
X^2+6X +2X+12 [12=6*2
8=6+2]
2:TAKE COMMON FACTORS FROM 1ST TWO AND
LAST TWO FACTORS
X^2+6X +2X+12 EQUALS
X(X+6)+2(X+6)
3:TAKE (X+6) AS COMMON FACTOR
X(X+6)+2(X+6) EQUALS

(X+6){X+2}


PS:WHILE SOLVING QUADRATIC EQNS IN THIS FORM WHERE X^2 HAS NO COEEFICIENT U CAN DIRECTLY WRITE THE SOLN AS (X+6)(X+2)

You have to use the sum and product rule for this one.

In an equation where y = x^2 + bx + c, you have to find a pair of numbers, so that:

#1 + #2 = b , and
#1 x #2 = c

so for this equation,

_ + _ = 8
_ x _ = 12

With trial and error, you will find the values 6 and 2

6 + 2 = 8
6 x 2 = 12

So now you can substitute these 2 numbers into the factorized equation.

y = (x + 6)(x+ 2)

(x+6)(x+2)

x^2+8x+12
x^2+6x+2x+12
x(x+6)+2(x+6)
(x+2)(x+6)

Look for a two numbers that
when you multiply equals 12, and
when you add make 8

for example, 6 and 2!

there fore it's (x + 6) (x + 2)
you can check by factoring.

x^2 + 8x + 12
= (x + 6)(x + 2)

(x + 6)(x + 2)
Check
x (x + 2) + 6 (x + 2)
x² + 2x + 6x + 12
x² + 8x + 12

1.using the factor method
x 6
x 2
2x +6x = 8x

= (x + 6)(x + 2)

The factors of Ax² + Bx + C are (x - Y) and (x - Z), where
Y = [-B + √(B² - 4AC)]/2A and Z = [-B - √(B² - 4AC)]/2A.

For x² + 8x + 12,
A = 1, B = 8 and C = 12.
This gives Y = -2 and Z = -6, so
x² + 8x + 12 = (x + 2)(x + 6)

get a ti-89