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qwertyuioop · Jun 14, 2012, 01:56 AM

how do you do this I did some of it but I'm not sure if I'm doing it correctly!
2x to the secomnd power +5x-7
I know that the answer will look something like this but how do u finish it? /
(2x___)(x___)
I need 2 now how 2 do it step by step!!
if you could please help me I would be really glad
because I have a MATH FINALL TODAY!!

push1t · Jun 14, 2012, 03:13 AM

(2x+7)(x-1)

qwertyuioop · Jun 14, 2012, 03:17 AM

How did u get that?

push1t · Jun 14, 2012, 03:22 AM

Well the way I look at it is that 2 numbers have to multiply together to make -7 and two numbers have to add together to make +5. it could only be 7 and 1 obviously altering the signs, since there's a 2 in front of the x you can multiply it by the one and you have 7. 7-2 gives 5. so you got your factors you just need to fiddle with the signs to see which is correct. Hope this helps.

push1t · Jun 14, 2012, 03:24 AM

Ignore the other one there's a slight error, here I corrected it Well the way I look at it is that 2 numbers have to multiply together to make -7 and two numbers have to add together to make +5. it could only be 7 and 1 obviously altering the signs, since there's a 2 in front of the x you can multiply it by the one and you have 2. 7-2 gives 5. so you got your factors you just need to fiddle with the signs to see which is correct. Hope this helps.

qwertyuioop · Jun 14, 2012, 03:24 AM

Omg thanks so much <3

qwertyuioop · Jun 14, 2012, 03:28 AM

can I ask u another one?
3x^2-5x^2+2

push1t · Jun 14, 2012, 03:43 AM

since 3x^2-5x^2 simplifies into -2x^2, you're left with -2x^2+2. it will be the form (-2x__)(x__) to fill in the blank spaces you need two number that multiply to form 2 yet add to form zero in this expression. It has to be 2 and 1. since -2*1 gives -2 and 2*1 gives 2, -2+2 gives zero, yet 2*1 gives 2. so the expression is (-2x+2)(x+1). The reason you need zero is because you assume the quadratic form ax^2 +bx +c, in the expression you gave a would be -2 b would be zero and c would be +2

qwertyuioop · Jun 14, 2012, 03:49 AM

I don't get that doesn't the # in the (2x__)(x__) have to = to 5
and how did u simplifi the 3 and y dose it need to be simplified and y is it negative?

qwertyuioop · Jun 14, 2012, 03:53 AM

Originally Posted by push1t

since 3x^2-5x^2 simplifies into -2x^2, you're left with -2x^2+2. it will be the form (-2x__)(x__) to fill in the blank spaces you need two number that multiply to form 2 yet add to form zero in this expression. it has to be 2 and 1. since -2*1 gives -2 and 2*1 gives 2, -2+2 gives zero, yet 2*1 gives 2. so the expression is (-2x+2)(x+1). The reason you need zero is because you assume the quadratic form ax^2 +bx +c, in the expression you gave a would be -2 b would be zero and c would be +2

I think the equatin that your doing is wrong its...
3x^2 -5x^2 +2

push1t · Jun 14, 2012, 06:51 AM

It's not because the first part of it can be simplified, since 3x^2 and -5x^2 have the x^2 term in common, you can reduce that bit to -2x^2. the answer I got was (-2x+2)(x+1) , if it wasn't negative than the answer would be (2x-2)(-x-1). What I'm trying to get as is there's more than one way to answer the question. The best way to check if you got the right answer is to expand the brackets and if you're right than you should end up where you started.