(X + 5)(X + 1)
FOIL this.
X2 + 1X + 5X + 5
Gather terms.
X2 + 6X + 5
----------------------------those were the factors
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
Factors are (x + 5)(x + 1) so x = -5 or -1
x3 - 7x2 + 6x = x(x2 - 7x + 6) = x(x2 - x - 6x + 6) = x[x(x - 1) - 6(x - 1)] = x2(x - 1) - 6x(x - 1) = (x2 - 6x)(x - 1)
(-3,-1)
x2 + 5x - 6x = x2 - x = x(x - 1) which is zero when x = 0 or x = 1
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
x2 + 6x + 5 can be factored into (x+1) (x+5)
Factors are (x + 5)(x + 1) so x = -5 or -1
x3 - 7x2 + 6x = x(x2 - 7x + 6) = x(x2 - x - 6x + 6) = x[x(x - 1) - 6(x - 1)] = x2(x - 1) - 6x(x - 1) = (x2 - 6x)(x - 1)
x2 - 6x + 5 = (x - 1)(x - 5)
(-3,-1)
x2+6x-7 = (x+7)(x-1) when factored
x2 + 5x - 6x = x2 - x = x(x - 1) which is zero when x = 0 or x = 1
A quadratic equation. If you wish to solve for x, you can do so as follows: -x2 + 6x + 7 = 0 x2 - 6x - 7 = 0 (x - 7)(x + 1) = 0 x ∈ {-1, 7}
That doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: -3 plus or minus i, where i is the square root of negative 1.
xx+3x+7=6x+18 xx+3x+7-(6x+18)=6x+18-(6x+18) xx-3x-11=0 Factors of -11: 1,-11 -1,11 Doesn't factor evenly, use quadratic
x2 + 6x = 7 ⇒ x2 + 6x + 9 = 7 + 9 ⇒ (x + 3)2 = 16 ⇒ x + 3 = ±4 ⇒ x = -7 or 1