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**nwtmike****Member**- Registered: 2013-06-06
- Posts: 1

For the quadratic 2X**2 + 7x + 3 please show the step in getting this 2x(x+3) + (x+3) = (2x+1)(x+3)

How did you get (2x+1)(x+3)?

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

2x^2+7x+3

=2x^2+6x+x+3

=2x(x+3)+1(x+3)

=(2x+1)(x+3)

There are 10 kinds of people in the world,people who understand binary and people who don't.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,647

hi nwtmike,

Welcome to the forum.

The trick is to find the right way to split up the 7x. In this case 6x and 1x. But how do you know that split? Why not 5x and 2x?

Look at the diagram below. It may help you to make the split but it still needs a bit of trial.

The four elements must be 2x^2, ?1, ?2 and 3. Put them into the boxes.

?1 and ?2 have to add up to 7x.

It is uncertain whether to put 1 or 3 with the 2x. Whichever you choose, the other will then be the other number, 3 or 1.

So try it out with 3 and 1 one way round, and if that fails, swap them around and you're done.

My second diagram shows what happens if you get the 3 and 1 the wrong way round.

Hope that helps with the next one.

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,647

hi gracejorn43

Thanks for the comment and welcome to the forum.

Bob

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**tutor****Member**- Registered: 2013-06-25
- Posts: 2

hi mike,bob explained well but i will give some explanation to understand you better

For factors first you have to know 4 formulas

+ * + = +

+ * - = -

- * + = -

- * - = +

Given equation be 2x^2+7x+3

quadratic equation be ax^2+bx+c=0 so here a=2, b=7, c=3

first multiply 2*3=6

now find the factors for 6

6=1*6

2*3

3*2

6*1

check according the formulas given above to get the value 7 because in given equation the value of B be 7

6*1=6 required factor if you add these you get 7 i.e.,6+1=7

if you add 2+3=5, 5 is the answer which is not the value of b. so take 6 and 1 as factors

+(6) *+(1)=+ (6) or +(1) *+(6)=+ (6)

2x^2+7x+3

2x^2+6x+1x+3

2x(x+3)+1(x+3)

(2x+1)(x+3)

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