I assume you are factoring. Don't worry, my students struggle with this too until they get a very critical idea, and we will get you to see it too!
I'll work through 2 examples with you, but I'm going to leave it to you to finish the others. just comment back if you are still stuck. deal?
ok. Lets look at the idea of factoring. the big idea is this, every part of an expression is made up with smaller parts, we just cant see them. these smaller parts are called FACTORS (which is why this is called factoring.
for example, the number 20 can be written a number of different ways using multiplication.
[tex]20 = 1 \times 20 \\ 20 = 2 \times 10 \\ 20 = 4 \times 5 \\ 20 = 5 \times 4 \\ 20 = 10 \times 2 \\ 20 = 20 \times 1[/tex]
we can do the same thing with negative numbers.
[tex] - 20 = - 1 \times 20 \\ - 20 = - 2 \times 10 \\ - 20 = - 4 \times 5 \\ - 20 = - 5 \times 4 \\ - 20 = - 10 \times 2 \\ - 20 = - 1 \times 20[/tex]
these numbers are called the FACTORS of 20.
we can do this with variables too!
[tex]x = 1 \times x \\ {x}^{2} = x \times x \\ {x}^{3} = x \times x \times x \\ {x}^{4} = x \times x \times x \times x[/tex]
and so on.
we can even do this with combinations (addition and subtraction) of numbers and variables.
for example
[tex]20 {x}^{2} + 30x[/tex]
can be rewritten as
[tex]1 \times 20 \times x \times x +1 \times 30 \times x \\ 2 \times 10 \times x \times x + 3 \times 10 \times x \\ 4 \times 5 \times x \times x + 5 \times 6 \times x \\ [/tex]
so, we need to find the greatest common factor of the numbers and greatest number of variables that go into each part of our expression. in our example the greatest common number is 10 and the greatest number of variables is one, since 30x only has one x as its factor.
now, remember that we cannot change the value of our expressions, so when we factor them out we need to keep it around, so we multiply it to what's left.
[tex]10x(2x + 3)[/tex]
when we have a situation like number 36, we can factor the binomial.
all factoring a binomial is really doing is undoing foil, or double distribution or box method or however expanding binomial was taught to you.
example
[tex](x + 4)(x + 3) = \\ x \times x + 3 \times x + 4 \times x + 4 \times 3 = \\ {x}^{2} + 7x + 12[/tex]
so, if we work this backwards we need to find two numbers that multiply to the last number and add to the middle number.
[tex] {x}^{2} + 7x + 12 = \\ {x}^{2} + 4x + 3x + 12[/tex]
now we pair them up with easy like terms.
[tex]( {x}^{2} + 4x) + (3x + 12)[/tex]
next, look at their factors to see what we can take out
[tex]( {x}^{2} + 4x) = \\ (x \times x + 4 \times 1 \times x) = \\ (x \times x + 2 \times 2 \times x )[/tex]
so we can only factor out an x out of our first parenthesis.
[tex]( {x}^{2} + 4x) = x(x + 4)[/tex]
the second parentheses can be worked the same way.
[tex](3x + 12) = \\ (3 \times 1 \times x + 12 \times 1) = \\ (3 \times x + 6\times 2) = \\ (3 \times x + 4 \times 3)[/tex]
they only have a 3 in common, so we will take that one out.
[tex](3x + 12) = 3(x + 4)[/tex]
so, now we have
[tex] {x}^{2} + 7x + 12 = \\ x(x + 4) + 3(x + 4)[/tex]
we have something else in common between these two pairs. do you see it?
[tex](x + 4)[/tex]
so, we factor that out
(x+4)(x+3)
look at 36.
[tex] \frac{1}{4} {x}^{3} [/tex]
has already been factored out for us, but the question wouldn't be there if we didn't have something else to do, so let's look at the parentheses.
[tex]( {x}^{2} - 9)[/tex]
do our factoring, and see if we can notice our factors.
what two numbers multiply to -9 and add to 0?
3 and -3, so we have 1/4×x^2(x×x+3x-3x+3×3)
pair them up and you get
(x×x-3x)+(3x-3×3)
x(x-3)+3(x-3)
(x-3)(x+3)
so our answer is
1/4×x^2(x-3)(x+3)
I'm going to work number 39 and 42 on paper
