I'm kinda banging my head for this one. Is this one has middle term?
3x^2+6x-3 ?
Help me plz.
tx
As Wilmer said, you factor out the "3" in each term: $\displaystyle 3(x^2+ 2x- 1)$. Other than that, it cannot be factored with integer coefficients. The fact is that [b]most[b] polynomials cannot be factored with integer coefficients.
(I keep saying "with integer coefficients" because we could say that $\displaystyle x^2+ 2x- 1= x^2+ 2x+ 1- 2= (x+1)^2- 2$. And now we can use the simple rule $\displaystyle a^2- b^2= (a- b)(a+ b)$ so that $\displaystyle x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2})$. Of course, to do that, I had to, effectively, solve the equation $\displaystyle x^2+ 2x- x= 0$ by 'completing the square' and, since a main application of factoring is to solve equations, we usually mean "with integer coeffcients". But, once again, most polynomial equations can't be factored that way.)
3x^2+6x-3
first you have to convert it into quadratic form of equation
i.e a x^2 + b x + c =0
then find D = b^2 - 4ac
if D>0 then factor of equation is possible
otherwise use
X1 = [ -b + sqrt(D) ] / 2a
X2 = [ -b - sqrt(D) ] / 2a
..... try it also helps you Quadratic Equation
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Thanks emakarov
I came across this formula, I heard this one is used when there is no easy way to factor. What is the story with this formula and how does it works?
$\displaystyle \frac{-b \pm\sqrt{b^2 - 4ac}}{2a} $
I'm a programmer and I am liking La Text
Edit: My bad, I fixed the sign before 4ac.