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No. They are the factors for x2 + 5x + 6.
Yes.
(3x + 5)(x2 - 2)
9x(x2+3x-10) * * * * * 9x3 + 27x2 - 90x = 9x(x2 + 3x - 10) = 9x(x + 5)(x - 2).
Multiply by 4. 4x2 - 12x + 9 factors to (2x - 3)(2x - 3) or (2x - 3)2
-x(x2 - 2)(x2 + 3)
x**2-9-9 factors into 3,-33+-3=0x**2-9=(x-3)(x+3)3x-9=3(x-3)(x-3)=LCM
(3x + 5)(x2 - 2)
(x + 1) and (x + 2) are monomial factors of the polynomial x2 + 3x + 2 (x + 1) and (x + 3) are monomial factors of the polynomial x2 + 4x + 3 (x + 1) is a common monomial factor of the polynomials x2 + 3x + 2 and x2 + 4x + 3
9x(x2+3x-10) * * * * * 9x3 + 27x2 - 90x = 9x(x2 + 3x - 10) = 9x(x + 5)(x - 2).
assuming x*2-3x+2:factors of -3:-1,3 --> sum: 21,-3 --> sum: -2(x-1)(x+3)
5
We have, f(x) = x2-3x-18 f(x) = x2-3x-18 = x2-6x+3x-18 = x(x-6)+3(x-6) = (x+3)(x-6) Factors are (x-6) and (x+3).
Multiply out the factors and then you can check: (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6 ≠ x² + 3x + 4 Therefore False.
x3 + x2 - 3x - 3 x(x2 + x - 3) - 3
x2/(3x2 - 5x - 2) - [2x/ (3x + 1)][1/(x - 2)] = x2/(3x2 - 6x + x - 2) - 2x/(3x + 1)(x - 2) = x2/[3x(x - 2) + (x - 2)] - 2x/(3x + 1)(x - 2) = x2/(3x + 1)(x - 2) - 2x/(3x + 1)(x - 2) = (x2 - 2x)/(3x + 1)(x - 2) = x(x - 2)/(3x + 1)(x - 2) = x/(3x + 1)
3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
X2 + 3X - 18 = 0What two factors of - 18 add up to 3 ??(X - 3)(X + 6)============so, by the zero sum rule,X = 3X = - 6
(3x - 3)(x/6)(x2 - x)= (3)(x - 1)(x/6)(x)(x - 1) = (1/2)(x2)(x -1)2 = (1/2)(x2)(x2 - 2x + 1) = (1/2)x4 - x3 + (1/2)x2