hi,
i need to factorise 2x^2-5x-3
i have been trying this for ages and just can get the correct answer so can someone help and point me in the right direction?
thanks!
If are the roots of then, .
Fernando Revilla
No, you shouldn't be looking at "1+ -3" at all! That would be true if there were no number multiplying either x: (x+ a)(x- b)= x^2- (b- a)x- ab so that you need "b- a" to be the number multiplying x. But (2x+ a)(x- b)= 2x(x)+ 2x(-b)+ a(x)+ a(-b)= x^2- (2b- a)x- ab= x^2- 5x- 3 if b= 3 and a= 1. Then -ab= -(1)(3)= -3 and 2b- a= 2(3)- a= 5.
andyboy179
you set up the first part correct . now you replace ? with two factors from the last term of the polynomial. the factors of 3 are 1,3. so try those you can use a simple rule to check if this is correct.
to check if your factors are correct
1) multiply the outer two terms together.
2) multiply the inner two terms together.
3) add the two numbers together. the sum should be the middle term in your polynimial.
so you would do this
1)
2)
3)
is the middle term of the polynomial so its correct.
you know the F.O.I.L rules?. you notice that what we did was use the O.I part of foil. if you keep going and do the F and L you expand the whole polynomial back to
Given ax^2 + bx + c
This is called the "ac" method. It looks a lot worse than it is, and the nice thing is that it will factor anything that can be factored (over the integers.)
Here a = 2, c = -3
Then ac = (2)(-3) = -6
The list of distict factors of -6 are:
6, -1
3, -2
2, -3
1, -6
Choose the pair that add up to b, in this case b = -5
So we have that 1 + -6 = -5
This means that you split the middle term of your quadratic in the following way:
-5x = (1)x + (-6)x
Thus
2x^2 - 5x -3 = 2x^2 + (x - 6x) - 3
= (2x^2 + x) + (-6x - 3)
Can you finish this from here?
-Dan
Hello, andyboy179!
i need to factorise:
i have been trying this for ages . . .
You said that you already have: .
The last term comes from four possible multiplications:
. .
Try all of them and see which gives a product of
I don't see how this can take "ages" . . .