Q: X2 plus 8x plus 16 equals?

Write your answer...

Submit

Still have questions?

Continue Learning about Other Math

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

2x^2 + 8x + 3 = 0

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

You ignore the constant part (-3 in this case), calculate half of the coefficient of the linear part (8, in this case), and square this half. (1/2 of 8 is 4; the square of 4 is 16). This gives you the perfect square x2 + 8x + 16. To get something equivalent to the original expression, you must both add and subtract 16, and include the term which I previously ignored (-3): x2 + 8x + 16 - 16 - 3, which you can write as (x2 + 8x + 16) - 16 - 3. The part within parentheses is the perfect square.

Why, yes, indeed, it has! Look at the factors: x2 - 8x + 16 = (x - 4)2 = (x - 4)(x - 4). If you multiply it out, you will see that it is so. Note that ' (x - 4) ' appears twice in the factorisation. That is what we mean, by a 'repeated factor'.

Related questions

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

2 this Domo

x*x - 8x +16 = 0 x*x = 8x + 16 x - 16 = _/8x x_/-4 = 2 x = 2

x2 + 8x - 6 = 0 x2 + 8x + 16 = 22 (x + 4)2 = 22 x + 4 = ± √22 x = -4 ± √22

x2+8x+9 = -7 x2+8x+9+7 = 0 x2+8x+16 = 0 (x+4)(x+4) = 0 Therefore: x = -4 and also x = -4 (they both have equal roots)

x2 + 8x - 5 = 0 ∴ x2 + 8x + 16 = 21 ∴ (x + 4)2 = 21 ∴ x + 4 = ± √21 ∴ x = -4 ± √21

no

In general, no.

x + 4

x2 + 8x - 2 = 0 => x2 + 8x = 2 => x2 + 8x + 16 = 2 + 16 = 18 => (x + 4)2 = 18 => x + 4 = +or- sqrt(18) = +or- 3*sqrt(2) so x = -4 +or- 3*sqrt(2)

x - x2 - 9x + 14 = 0 ; whence, x2 + 8x = 14 , x2 + 8x + 16 = 30 , and x + 4 = ±√30 . Therefore, x = ±√30 - 4 .

x2 - 8x = 1 ∴ x2 - 8x + 16 = 17 ∴ (x - 4)2 = 17 ∴ x - 4 = ±√17 ∴ x = 4 ± √17