factor the trinomial?

Question

x2power + 5x + 6


x 2power + 6x + 5

x2power + 3x + 2

x2power - 3x + 2

x2power - 7x + 12

x2power - 5x + 6

x2power - x - 2

x2power - 4 x - 12

x2power - 2x - 8

Answer 1

x2power + 5x + 6
Think of 2 numbers that when added equal 5, and when multiplied equal 6. How about 3 and 2?

= (x+3)(x+2)


x 2power + 6x + 5 = (x+1)(x+5)

x2power + 3x + 2 = (x+2)(x+1)

x2power - 3x + 2 = (x-2)(x-1)

x2power - 7x + 12 = (x -3)(x-4)

x2power - 5x + 6 = (x-3)(x-2)

x2power - x - 2 = (x-2)(x+1)

x2power - 4 x - 12 = (x-6)(x+2)

x2power - 2x - 8 = (x-4)(x+2)

Answer 2

The objective for factoring is finding two numbers that when multiplied together give you the last term (meaning two factors of the last number) and when added give you the middle term

x2power + 5x + 6
(x+3)(x+2)


x 2power + 6x + 5
(x+1)(x+5)
x2power + 3x + 2
(x+1)(x+2)
x2power - 3x + 2
(x-1)(x-2)
x2power - 7x + 12
(x-3)(x-4)
x2power - 5x + 6
(x-3)(x-2)
x2power - x - 2
(x+1)(x-2)
x2power - 4 x - 12
(x-6)(x+2)
x2power - 2x - 8
(x-4)(x+2)

Answer 3

x² + 5x + 6 = 0

x² + 3x + 2x + 6 = 0

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

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

- - - - - - - - - - -s-

Answer 4

(x+3)(x+2)
(x+5)(x+1)
(x+2)(x+1)
(x-2)(x-1)
(x-4)(x-3)
(x-2)(x-3)
(x-2)(x+1)
(x-6)(x+2)
(x-4)(x+2)