ok, whilst factorizing polynomials of this way, a million elementary rule helps. seem for 2 numbers that upload to supply the middle term and multiply to supply the third term thus that's 2 and 3 and your factorized eqn is as such: (x+2)(x+3) desire this helps
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Verified answer
Assume ax^2 + bx +c = 0.
a = 1
b= 1
c = -6
To find the factors we must assume x1* x2 = ac and x1+x2 = b
In this case 3 * -2 = -6 (a*c)
and
3 + -2 = 1 (b)
Now we take -2 and 3 and set it as a linear factor.
To make sure that we get the right factor simply say that
(x+2) = 0 x =-2
(x-3) = 0 x = 3
So our factors are (x+2)(x-3)
(x+2)(x-3) = 0
To check:
FOIL
First: x(x) = x^2
Outer: x(-3) = -3x
Inner: x(2) = 2x
Last : 2(-3) = -6
x^2 -3x + 2x -6 combine like terms:
x^2 + (-3+2)x -6
x^2 + x - 6 = 0
It checks out :)
Hope this helps!
x²+x-6 = (x+3)(x-2)
ok, whilst factorizing polynomials of this way, a million elementary rule helps. seem for 2 numbers that upload to supply the middle term and multiply to supply the third term thus that's 2 and 3 and your factorized eqn is as such: (x+2)(x+3) desire this helps
I assume that you meant
x^2 + x - 6 = (x+3)(x-2)
x^2 + x – 6
Factors of -6 that add to +1 are -2 & +3
x^2 + 1x - 6 = (x - 2)(x + 3)
(x + 3)(x - 2)
(x+3)*(x-2)
It equals (x^2+3x)+(-2x-6),which when you combine like terms is the same as the original equation.