(x2+3x-4) (x-4)
3x3 - x2y + 12x - 4y = x2*(3x - y) + 4*(3x - y) = (x2 + 4)*(3x - y)
-2x + 3x = -4 By combining like terms we get: x = -4.
Hmm - you need to phrase your question a little better. Which of these are you looking for? x2 - 3x + 2x - 2 = x2 - x - 2 = (x - 2)(x + 1) x2 - (3x + 2)(x - 2) = x2 - 3x2 + 6x - 2x + 4 = -2x2 + 4x + 4 = -2(x2 - 2x - 2) x2 - 3x + 2(x - 2) = x2 + x - 4
3x-1 = 11 3x = 11+1 3x = 12 x = 4 Therefore: x2+2 = 18
The discriminant of x2+3x+4 is -7 therfore it has no real roots.
(x2+3x-4) (x-4)
(x2 + 3x + 5)(x2 - 3x + 4)
It is difficult to tell because there is no sign (+ or -) before the 5. +5 gives complex roots and assuming that someone who asked this question has not yet come across complex numbers, I assume the polynomial is x2 -3x - 5 The roots of this equation are: -1.1926 and 4.1926 (to 4 dp)
For any x ≠ 0, x2 -10/x2 - 4 +3x/x2 - 4 LCD = x2, multiply each term by their missing element of LCD = (x4 + 10 +3x - 8x2)/x = (x4 - 8x2 + 3x + 10)/x2
3x3 - x2y + 12x - 4y = x2*(3x - y) + 4*(3x - y) = (x2 + 4)*(3x - y)
y = x2-4x+4 Since the highest degree term is 2, it must have 2 roots
x2 + 3x + 28 can not be factored. If on the other hand you meant to type: x2 + 3x - 28 That can indeed be factored: = x2 + 7x - 4x - 28 = x(x + 7) - 4(x + 7) = (x - 4)(x + 7)
(3x4 + 2x3 - x2 - x - 6)/(x2 + 1)= 3x2 + 2x - 4 + (-3x - 2)/(x2 + 1)= 3x2 + 2x - 4 - (3x + 2)/(x2 + 1)where the quotient is 3x2 + 2x - 4 and the remainder is -(3x + 2).
-2x + 3x = -4 By combining like terms we get: x = -4.
Hmm - you need to phrase your question a little better. Which of these are you looking for? x2 - 3x + 2x - 2 = x2 - x - 2 = (x - 2)(x + 1) x2 - (3x + 2)(x - 2) = x2 - 3x2 + 6x - 2x + 4 = -2x2 + 4x + 4 = -2(x2 - 2x - 2) x2 - 3x + 2(x - 2) = x2 + x - 4
3x3 + 6x2 - 24x = 3x(x2 + 2x - 8) = 3x(x2 + 4x - 2x - 8) = 3x[ x(x + 4) - 2(x + 4) ] = 3x(x - 2)(x + 4)