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Q: Factor x2 plus 8x plus 12?

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

x2 + 8x + 15 = (x + 3)(x + 5)

x2 + 8x + 15 = (x + 3) (x + 5).

That does not factor neatly.

x2 + 8x = x(x+8)

8x(x2-x+1)

x + 4

x2-8x+15 = (x-3)(x-5)

x2 - 8x + 12 (x - 6)(x - 2) x2 - 2x - 6x + 12 x2 - 8x + 12 x - 6 = 0 x = 6 x - 2 = 0 x = 2 Solution set: {2, 6}

x2-8x+12 = (x-2)(x-6) when factored

x2-8x+16 = (x-4)(x-4) when factored

(x + 6)(x + 2)

(x+1)(x+7)

2x(x2+2x+4)

No because the discriminant is less than zero.

x2 - 8x + 32 = 0 (assuming)you can factor or use the quadratic equation.I recommend the quadratic equation.

x - 12

X2 + 8x + 16 = 10x +16x2 + 8x + 16=2x + 8x + 16=10x + 16

(x-4)(x-4)

3(x2-x-7)

# (x ) ( x ) # ## (x + 2 ) ( x + 6 )

4x2 + 8x + 4 = 4 (x2 + 2x + 1) = 4 (x + 1)2

-6

If you notice, 8x + 4 = 4(2x + 1), but there is an odd coefficient for x2, so there is guaranteed to be some remainder (that is 8x + 4 is not a factor of the polynomial): (24x3 - 5x2 - 48x - 8) ÷ (8x + 4) = 3x2 - 2x - 5 - (x2 - 12)/(8x + 4)

x2 + 8x - 20 = (x - 2)(x + 10).