Factorise.. x^2 - y^2 - 2x + 6y - 8?
Hi, instead of -8 write -9 + 1 x^2 - y^2 - 2x + 6y - 9 + 1 = x^2 - 2x + 1 - y^2 + 6x - 9 = x^2 - 2x + 1 - (y^2 - 6x + 9) = (x - 1)^2 - (y - 3)^2 = (x-1 + y-3)(x-1 - (y-3)) = (x + y - 4)(x - y + 2)
How do you factor x² -x -6 =0?
(x-3)(x+2) since there is a minus in front of the 6 the signs in front of the factors are different. the two factors have to multiply to get the third term. -3*2 = -6 the two factors have to add up to get the second term. -3+2=-1 or -x
How do you factor x^3+2x^2-5x-6?
For cubics and above, unless you remember the formula you are best using trial and error to guess at some possible solutions. For example 0 and 1 are not solutions but -1 is. Now divide the cubic by x+1 (x^3 + 2x^2 - 5x - 6)/(x + 1) = x^2 + x - 6 Now for this quadratic you can use the quadratic formula but with practice you'll immediately see that x^2 + x - 6 = (x + 3)(x - 2) so... x^3 + 2x^2 - 5x - 6 = (x + 1)(x - 2)(x + 3)
How to factor 12x^2 + x - 6?
The most straight-forward way is to use the quadratic formula. The quad. formula finds the solutions to equations like a*x^2+b*x+c=0. in this case, ( -b+-sqrt(b^2-4ac) )/a gives 2/3 and -3/4 if you write that in the form of (x-a)*(x-b) you get (x-2/3)*(x+3/4) multiplying the whole thing by 12 to cancel out the denominators, you get (3x-2)*(4x+3) foiling gives 12x^2+9x-8x-6, which verifies my answer
How do you factor 6x^2+x-2?
(2x-1)(3x+2) You want the outsides multiplied added to the insides multiplied to equal 1 (since 1 is the coefficient of x) and -1 and 2 multiplied equal -2. 2 and 3 equal 6 (the coefficient of 6x squared)
Factor: y^2 - 100 ( y - 10)^2 2. ( y - 10)( y - 10) 3. ( y + 10)( y + 10) 4. ( y + 10)( y - 10)?
1. Factor y^2 - 100: > (y - 10) (y + 10) 2. Factor x^2 - x - 6 > (x - 3) (x + 2) 3. Solve the system of equations: Rewrite the second equation in terms of x: > x = 2 - y Substitute the new x into the other equation: > 4x + y = 11 > 4(2 - y) + y = 11 > 8 - 4y + y = 11 > -3y = 3 > y = -1 Substitute the value for y into the second equation and solve for x: > x + (-1) = 2 > x = 3
Factorize x³-6x² +3x+ 10?
Apply trial and error to the equation by substituting x as any random integer to the equation [math] x^3 - 6x^2 + 3x + 10 = 0 [/math]. If the equation gives a result of 0, then (x - root) will be one factor of the equation. In that case, consider x = 2, so [math] 2^3 - 6(2)^2 + 3 (2) + 10 = 0 [/math] [math] 8 - 24 + 6 + 10 = 0 [/math] Hence, (x - 2) is one of the roots of the equation. Now, apply long division, and accordingly we will obtain another factor as [math] x^2 - 4x - 5 = (x - 5) (x + 1) [/math]. Hence, [math] x^3 - 6x^2 + 3x + 10 = (x + 1) (x - 2) (x - 5) [/math].
Can i factorise this equation x^3 - 7x - 6?
x^3 - 7x - 6 = x^3 - 2x^2 - 3x + 2x^2 - 4x - 6 = (x^3 - 2x^2 - 3x) + (2x^2 - 4x - 6) = x(x^2 - 2x - 3) + 2(x^2 - 2x - 3) = (x^2 - 2x - 3)(x + 2) = (x^2 + x - 3x - 3)(x + 2) = [(x^2 + x) - (3x + 3)](x + 2) = [x(x + 1) - 3(x + 1)](x + 2) = (x + 1)(x - 3)(x + 2)