# Math Help - middle term / factor question..

1. ## middle term / factor question..

I'm kinda banging my head for this one. Is this one has middle term?

3x^2+6x-3 ?

Help me plz.

tx

2. ## Re: middle term / factor question..

Well, divide by 3 to start: x^2 + 2x - 1
If it was x^2 + 2x + 1, then that would factor to (x + 1)^2

3. ## Re: middle term / factor question..

As Wilmer said, you factor out the "3" in each term: $3(x^2+ 2x- 1)$. 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 $x^2+ 2x- 1= x^2+ 2x+ 1- 2= (x+1)^2- 2$. And now we can use the simple rule $a^2- b^2= (a- b)(a+ b)$ so that $x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2})$. Of course, to do that, I had to, effectively, solve the equation $x^2+ 2x- x= 0$ 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.)

4. ## Re: middle term / factor question..

thanks,

Hei HallsofIvy

How do you write maths characters like that?

5. ## Re: middle term / factor question..

Originally Posted by ameerulislam
How do you write maths characters like that?
You have to use LaTeX coding.
$$x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2})$$ gives $x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2})$.

6. ## Re: middle term / factor question..

Originally Posted by Plato
You have to use LaTeX coding.
$$x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2})$$ gives $x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2})$.
where can I get codes for all the characters I'm looking for. I want the code for under line, I mean 2/3 .. Two over three here.... and code of the line between them.

7. ## Re: middle term / factor question..

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

8. ## Re: middle term / factor question..

Originally Posted by ameerulislam
where can I get codes for all the characters I'm looking for.
See the LaTeX Help subforum. Remember to use $$...$$ tags. See also the LaTeX Wikibook or search the web.

9. ## Re: middle term / factor question..

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?

$\frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$

I'm a programmer and I am liking La Text

Edit: My bad, I fixed the sign before 4ac.

10. ## Re: middle term / factor question..

Originally Posted by ameerulislam
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 work?

$\frac{-b \pm\sqrt{b^2 + 4ac}}{2a}$
The expression under the square root (called discriminant) should be $b^2-4ac$. I personally use this formula instead of factoring. It gives the two roots of the quadratic equation $ax^2+bx+c=0$.

11. ## Re: middle term / factor question..

I have this equation to factor
$x^2 - 4x +5$

$\frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$

within square root it comes -16.. So does that means I can't factor that equation?

12. ## Re: middle term / factor question..

Originally Posted by ameerulislam
I have this equation to factor
$x^2 - 4x +5$ $\frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$
within square root it comes -16.. So does that means I can't factor that equation?
Actually it is $\sqrt{-4}$. Using complex numbers:
$(x-2+i)(x-2-i)$

13. ## Re: middle term / factor question..

yeah only complex number will fit there. I think I can ignore it as I'm just doing simple calculus (only real numbers)..

14. ## Re: middle term / factor question..

Originally Posted by ameerulislam
yeah only complex number will fit there. I think I can ignore it as I'm just doing simple calculus (only real numbers)..
Well then, why isn't the expression x^2 - 4x - 5 ?