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I'm dong some Factorisation revision and I'm a bit stuck. I've never done this before and I'm stuck a bit. First of all I know how to do small factorisations and I know you have to find the Highest Common Factor and that it's the opposite of expanding but these expressions are scaring me a bit.

e.g. how would you do this $x^2 + 8x + 16$

or this $9m^2 - 24mn + 16n^2$

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@IanMateus, it was tagged as homework. You should only give some hints instead of solving it. –  Sigur May 30 '13 at 15:54
    
@Sigur you are right. Deleting it. –  Ian Mateus May 30 '13 at 15:55
    
OK. But now I saw that the OP asked other questions like this and there are many complete answers. So they already gave the solution to her. –  Sigur May 30 '13 at 15:56
    
Any hints then??? I didn't get the solved one lol. –  user61406 May 30 '13 at 15:57

6 Answers 6

Remember, and you must learn it by heart: $$(a{+}b)^2=a^2+2ab+b^2$$ and $$(a-b)^2=a^2-2ab+b^2$$ So try to write each of two expressions in one of this two forms

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"I was made to learn by heart: ‘The square of the sum of two numbers is equal to the sum of their squares increased by twice their product.’ I had not the vaguest idea what this meant and when I could not remember the words, my tutor threw the book at my head, which did not stimulate my intellect in any way." - Bertrand Russell –  Ian Mateus May 30 '13 at 16:04
    
I don't mean learn by heart learn without understanding but the memorization side in learning mathematics is essential. Isnt'it? –  Sami Ben Romdhane May 30 '13 at 16:14
    
$(a - b)^2 = 9m^2 + 16n^2 - 24mn$... what do I do now??? ? $(9m^2 + 16n^2)^2$ ??? You're telling me to put into one of those forms and that's what I did, lol I think it's wrong –  user61406 May 30 '13 at 16:22
    
@HonkyHanka The term $9m^2$ should have the form $a^2$ so $a=3m$ and $16n^2$ is the term $b^2$ so $b=?$ –  Sami Ben Romdhane May 30 '13 at 16:29
    
$b = 4n$ @SamiBenRomdhane but how would you do $x^2 + 8x + 16$? I know you have to put it into forms $(a + b)^2 = a^2 + 2ab + b^2$ but I don't know where to put the terms as there's only one squared and then there's $8x$ is that $a$ $b$ or $2ab$ –  user61406 May 30 '13 at 16:42

To solve these kinds of exercises all you need to know is $$ (a+b)^2=a^2+b^2+2ab,\quad (a-b)^2=a^2+b^2-2ab $$ and then be a little smart.

Suppose we are looking at $9m^2-24mn+16n^2$ and want to write it in one of the above forms. Since there is a minus-sign involved, we should bring it on the form $(a-b)^2$ for some $a$ and $b$ to be determined. Now $9m^2$ should correspond to the $a^2$ term, $16n^2$ should correspond to the $b^2$ term and $-24mn$ corresponds to $-2ab$.


Edit upon OP's now deleted comment: Aligning your expression with the expression we obtain from expanding $(a+b)^2$ we have: $$ \begin{align} &a^2\;\;\;+b^2\;\;\;\,-2ab\\ =&9m^2+16n^2-24mn \end{align} $$ meaning that you should pick $a$ and $b$ such that $a^2=9m^2$, $b^2=16n^2$ and $-2ab=-24mn$. Which $a$ and $b$ does this correspond to?

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I know I got that but what do you mean which $a$ and $b$ does it correspond to? –  user61406 May 30 '13 at 16:27
    
@HonkyHanka: I mean: You know that $(a-b)^2=a^2+b^2-2ab$ for all $a$ and $b$. Now, you have an expression ($9m^2+16n^2-24mn$) which is almost of the form $a^2+b^2-2ab$. And you know that if you can bring your expression on this form for some $a$ and $b$ (of course these will involve $m$ and $n$), then you're done. For example: I want $9m^2$ to correspond to the $a^2$ term. This is clearly satisfied if I put $a=3m$ since $(3m)^2=9m^2$, right? Now what should you choose $b$ to be? –  Stefan Hansen May 30 '13 at 16:32
    
$a = 3rn$ and $b = 4n$ –  user61406 May 30 '13 at 16:43
    
@HonkyHanka: Exactly. I take you mean $3m$ and not $3rn$. –  Stefan Hansen May 30 '13 at 17:23

You need to find the factors of the term with the power of 2 and the last term (which can also be in the power of 2)

Let's take this one:

$4a^2 - 16ab - 9b^2$

We know that the factors of the coefficient of the term in $a^2$ are 1, 2 and 4 (we can take the all positives here and ignore the negative factors to make it easier) and for the coefficient of the term in $b^2$, they are 1, 3, 9, -1, -3 and -9.

My method (there are several) is to put the brackets and the variables:

$$(a\ \ \ b)(a\ \ \ b)$$

I try first with 1 and 4 for $a^2$:

$$(a\ \ \ b)(4a\ \ \ b)$$

                                    1  2 3  4

Now, the only possible terms in $ab$ will be $4ab$ (#2 times #3) and $ab$ (#1 times #4), but we need to get $-16ab$ (see original expression), so that we need $-20ab$ to add to the current $4ab$. How can we get $-20$ from the factors 1, 3, 9, -1, -3 and -9, and $4ab$? Doesn't seem possible. So maybe putting #2 as $3b$, meaning #4 has to be $3b$ as well...

$$(a\ \ \ 3b)(4a\ \ \ 3b)$$

But this gives $12ab+3ab$, or if we take the negative, $-12ab-3ab$ not very useful... so probably 1 and 4 are not good. Let's try 2 and 2.

$$(2a\ \ \ b)(2a\ \ \ b)$$

                                    1  2 3  4

Now, the only possible terms in $ab$ will be $2ab$ (#2 times #3) and $2ab$ (#1 times #4), but again we need to get $-16ab$, so that we need $-20ab$ to add to the current $4ab$. How can we get $-20$ from the factors 1, 3, 9, -1, -3 and -9, and $2ab$, $2ab$? Well, here, you can notice we can have $-9\times2ab + 1\times 2ab = -16ab$. So that:

$$(2a-9b)(2a+b)$$

If we expand it, we get: $4a^2 - 18ab + 2ab -9b^2 = 4a^2 - 16ab - 9b^2$

It takes time and a lot of practice to make all the process faster (especially the part where you have to see what numbers go where) and more natural, but if you keep at it, it will remain in your head for a long time!

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you are not getting other people are saying people are using this formula

$$(a-b)^2=a^2+b^2-2ab$$

means you have to make terms in square form like in case of

$9m^2 - 24mn + 16n^2\implies(3m)^2+(4n)^2-2\times3m\times 4n\implies (3m-4n)^2$

try yourself first one.And one more thing bracket do not affect this question

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Given:

$\boxed{x^2 + 8x + 16}$

The first question is best solved by guessing. We need to find a number that multiplies to positive $16$ and adds to positive $8$.

Two numbers which fit this are: positive $4$ and positive $4$. Because $\sqrt{16}$ = $4$, we can call this a perfect square.

Therefore, we can factor this as: $$(x+4)(x+4)$$

Which is the same as: $$(x+4)^2$$

We can check this by using the FOIL method to expand the factoring:

$$x^2+4x+4x+16=x^2+8x+16$$

The next question is a little more difficult and so we will make use of a slightly different strategy.

Given:

$\boxed{9m^2 - 24mn + 16n^2}$

We can use a method called grouping to factor this equation.

We start by multiping $9m^2\cdot16n^2$ which equals $144m^2n^2$. Similarily to the last question we want to find two numbers which multiply to $144m^2n^2$ and add up to $-24mn$.

In this case, $-12mn$ and $-12mn$ are two numbers which multiply to positive $144m^2n^2$ and add to $-24mn.$ We now want to split the $-24mn$ and use grouping to factor this.

$$(9m^2-12mn)(-12mn+16n^2)$$

Notice how in both cases we can factor $3m$ out:

$$3m(9m^2-12mn)-3m(-12mn+16n^2)$$

Leaving us with:

$$3m(3m-4n)-4n(3m-4n)$$

We now have our final factored form:

$$(3m-4n) (3m-4n)$$

Which can also be written as:

$$(3m-4n)^2$$

This is also a perfect square.

Once again, we can check this is the correct factoring by simply expanding it using the FOIL method.

$$(3m-4n) (3m-4n) = 9m^2-12mn-12mn+16n^2\\=9m^2-24mn+16n^2$$

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I am going go try and answer my question. So you have to use the forms right: $(a + b)^2 = a^2 + 2ab + b^2$

and $(a − b)^2 = a^2 − 2ab + b^2)$

So for $9m^2−24mn+16n^2$ I would use the the first form right? So $(a + b)^2 = 9m^2 + 16n^2 + 24mn$

So $(3m^2 + 4n^2)^2$ as $3^2 = 9$ and $4^2 = 16$

$(a + b)^2 = x^2 + 16 = 8x$

So $(x^2 + 16)^2$

It's $(x + 4)^2$ as $x$ x $x = x^2$ and $4$ x $4$ $= 16$ with the $^2$.

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check the whole square terms those are $9m^2$ and $16n^2$ they will be $(3m)^2$ and $(4n)^2$.so according to formula $a=3m$ and $b=4n$ .and you have two forms so decide the form basis of sign of term "2ab".in ques this term is $-24mn$ so choose $(a-b)^2$ –  iostream007 May 30 '13 at 16:44
    
for second question 16 is also $4^2$ and another square is $x^2$ so you just need to break terms $8x$ –  iostream007 May 30 '13 at 16:49
    
Oh ok I got it now, I think is it right? –  user61406 May 30 '13 at 16:50

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