x2-12x+35 = (x-5)(x-7)
(12x - 6)(12x + 6)
2(3x^2 + 6x + 2)
(x - 8)(x - 4)
2x(x^2+x-6)
9x2 + 12x + 4 = 9x2 + 6x + 6x + 4 = 3x(3x + 2) + 2(3x + 2) = (3x + 2)(3x + 2) = (3x + 2)2
(12x - 6)(12x + 6)
No. It's a binomial, (whose value happens to be zero no matter what 'x' is).
(x - 3)(x - 9)
3x3+12x = 3x(x2+4)
-2x+12x+28=10x+28 ;Combining like factors2(5x+14) ; Factor out a two from both of them
30+12x = 6(5+2x)
x2 + 12x + 32 = (x + 8)(x + 4)
x2 + 12x + 32 = (x + 4)(x + 8)
(x + 4)(x + 8)
(x + 10)(x + 2)
To factor the trinomial ( x^2 - 12x + 35 ), we need to find two numbers that multiply to 35 and add to -12. The numbers -5 and -7 satisfy these conditions. Therefore, the trinomial can be factored as ( (x - 5)(x - 7) ), and one of the binomial factors is ( x - 5 ).
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: -6 plus or minus (3 times the square root of 2) x = -1.75735931288 x = -10.24264