Math Is Fun Forum

Ironsoul

Member

Registered: 2011-02-15

Posts: 12

Quadratic Equations: Solving by Factorisation Problems.

Hey again, I have a few more questions that the answer is not immediately clear to me.

Starting off: (note I'm still learning how to write in that math mode, so bear with me).

Steps showing how the answer is achieved will be greatly appreciated.

Edit: Another Question that's got me.

And another:

This style of question just gets me every time:

I've done this type before, but I didn't really get it then:

one more for the "need help" pile:

which i then turn into:

and this is where it gets confusing.

Last edited by Ironsoul (2011-02-20 14:52:45)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Quadratic Equations: Solving by Factorisation Problems.

Hi Ironsoul;

If you remember your FOIL then you know you need

( b +- "something" )(b +- "something else")

Whatever that "something" is, it has to sum to -1 (the middle term) and when multiplied to "something else" it has to equal the -2 ( the last term)

Can you solve from there? Where are you having problems?

Now you try this one:

You do this one the same way.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,142

Re: Quadratic Equations: Solving by Factorisation Problems.

hi Ironsoul,

When I teach classes to multiply out a pair of brackets leading to a quadratic I use a diagram like the one below.

Then, when I move on to factorising quadratics I use the diagram again, but working in reverse.

Take your first one:

(i) I put the b squared and -2 terms into the top left and bottom right spaces on the diagram.  I also put bs (shown in red) in the top space and left space to show that b x b = b squared.

(ii)  I look for two numbers that multiply to give -2 and also add or subtract to give the -1 in -1b.  With bigger numbers it may take a little while to find the right numbers.  In this case 2 and 1 will work.  And as I want -1b I know that -2 and + 1 are the correct signs for these.  So I add these  to the top and left on the diagram.

(iii)  Finally fill in the two 'b' spaces, checking that the middle term of my quadratic has come out correctly.

Then the top of the diagram (b - 2) and the left side (b + 1) give me the factorisation that I want.

If you go to www.bundy.demon.co.uk and click the lessons link you will find a whole load of lesson plans that I made for various classes.  There is one called 'quadratic expressions' which will give you more examples (it's a WORD.doc).  Sometime I will have to edit this sheet as it refers to the pages of a text book which won't mean anything to you I expect. Sorry.

Anyway, I hope that helps. 

Bob

Last edited by Bob (2011-02-20 21:49:03)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

4DLiVing

Member

Registered: 2011-01-08

Posts: 22

Re: Quadratic Equations: Solving by Factorisation Problems.

Hey Ironsoul
For problems like these, I always liked this method. Its similar to bob bundy's method, and will show you more steps.

3x² + 7x + 4.

1. Multiply the outside numbers 3 * 4 = 12

2. Find two factors of 12 that add up to 7.

In this case, the two factors of 12 that add up to 7 are: 3 and 4 or;

3 + 4 = 7 and

3 * 4 = 12.

Rewrite your original equation replacing the middle term

7x

with

3x + 4x, since 3x + 4x = 7x.

So instead of...

3x² +     7x      + 4, we now have...

3x² + 3x + 4x + 4.

factor out the 3x from first two terms, and the 4 from the second two...

3x(x + 1) + 4(x + 1).

Now pull out the (x + 1) and you have....

(x + 1)(3x + 4)