# Math Help - Homework help, please?

In my Algebra class, we are learning about factoring. I really don't understand most of it and was wondering if someone could help and explain the problems.

The directions are to solve each equation.(The answers can be whole numbers or fractions.)

1.) (3x-7)(x+5)=0.
2.) w^2=16w.
3.) B^2+6b-16=0.
4.) x^2-6x-27=o.
5.) y^2+23y=24.
6.) y^2-28=3y.
7.) 3x(2x+1)=o.

Any help would be great. Thanks.

2. Well, for number 1:

$(3x-7)(x+5)$

Have you heard of something known as F.O.I.L? well, this stands for first, outer, inner, last. This is what you would use to solve a problem such as this. To start, you multiply both of the FIRST terms in the parenthesis, which are 3x and x. You will recieve $3x^2$. next, you multiply the first and the OUTER (notice, the outer is the last one), these are 3x and 5. This is $15x$. Then you multiply both of the INNER terms, which are -7 and x, and get $-7x$. Then you multiply both of the LAST(last in each of the parentheses) terms, which are -7 and 5 and get $-35$. After this, you combine like terms.

$3x^2+15x-7x-35$

$3x^2+8x-35$

And there we go this factors like: $(3x-7)(x-5)$ set them equal to zero, and solve for x. and there is your answer.

for number 5:
$y^2+23y-24=0$

ok, you first set everything equal to zero, as i did above, then you look at the last term, and think: are there any factors of -24 that will add up to 23. YES there are these two factors will be 24 and -1. so, to factor this:

$(y-1)(y+24)$

like so.
you just plug in the factors. Note that this will only work when the coefficient of the first variable is 1 (see how $y^2$ is by itself? this will only work if $y^2$ has no coefficient.

Now you set y-1 equal to zero. So, y-1=0, and y=1. you do the same for y+24. Y+24=0 Y=-24. So, Y=1 and Y=-24.
for number 7: you will use the distributive property.

Problem: $3x(2x+1)=0.$ Note, if you set 2x+1 equal to zero, and solve, you will get -1/2 this is also a solution.

You will distribute the number outside of the parenthesis to the numbers in the parenthesis. This means you will multiply 3x by 2x and by 1. so you get:

$6x^2+3x=0$

thus, x=0
x=-1/2

3. Originally Posted by bigcitydreams
1.) (3x-7)(x+5)=0.
If you multiply two numbers and get zero, then you know that one (or both) of the numbers is itself zero. In this case, your factors are $3x-7$ and $x+5.$ Any value of $x$ that makes either one of these zero will satisfy the original equation. So split the problem up and solve, separately:

$3x-7=0$ and $x+5=0.$

2.) w^2=16w.
$\Rightarrow w^2-16w=0.$

Look on the left. Is there a common factor you can pull out of the expression?

3.) B^2+6b-16=0.
Did you mean to type the same letter for both variables?

If so, then to factor the expression you need to find two numbers that give 6 when added together and -16 when multiplied together. -16 can be factored as follows:

$\begin{array}{rr}
16\cdot-1&-16\cdot1\\
2\cdot-8&-2\cdot8\\
4\cdot-4&
\end{array}$

Now which pair of numbers above adds up to 6?

4.) x^2-6x-27=o.
-27 factors as

$\begin{array}{rr}
27\cdot-1&-27\cdot1\\
9\cdot-3&-9\cdot3\\
\end{array}$

Can you do it?

5.) y^2+23y=24.
$\Rightarrow y^2+23y-24=0.$

Use the same method as above.

6.) y^2-28=3y.
$\Rightarrow y^2-3y-28=0.$

Go on.

7.) 3x(2x+1)=o.
Set each factor equal to zero,

$3x=0$ and $2x+1=0.$

4. Originally Posted by rtblue
Well, for number 1:

$(3x-7)(x+5)$

Have you heard of something known as F.O.I.L? well, this stands for first, outer, inner, last. This is what you would use to solve a problem such as this. To start, you multiply both of the FIRST terms in the parenthesis, which are 3x and x. You will recieve $3x^2$. next, you multiply the first and the OUTER (notice, the outer is the last one), these are 3x and 5. This is $15x$. Then you multiply both of the INNER terms, which are -7 and x, and get $-7x$. Then you multiply both of the LAST(last in each of the parentheses) terms, which are -7 and 5 and get $-35$. After this, you combine like terms.
The problem was to solve the equation, not multiply the expression on the left.

$3x^2+15x-7x-35$

$3x^2+8x-35$

And there we go this factors like: $(3x-7)(x-5)$ set them equal to zero, and solve for x. and there is your answer.
NO, it does not factor like that: $(3x- 7)(x- 5)= 3x^2- 7x- 3x+ 35= 3x^2-10x+ 35$. The factors are exactly what was originally given (3x-7)(x+ 5)!

for number 5:
$y^2+23y-24=0$

ok, you first set everything equal to zero, as i did above, then you look at the last term, and think: are there any factors of -24 that will add up to 23. YES there are these two factors will be 24 and -1. so, to factor this:

$(y-1)(y+24)$

like so.
you just plug in the factors. Note that this will only work when the coefficient of the first variable is 1 (see how $y^2$ is by itself? this will only work if $y^2$ has no coefficient.

Now you set y-1 equal to zero. So, y-1=0, and y=1. you do the same for y+24. Y+24=0 Y=-24. So, Y=1 and Y=-24.
for number 7: you will use the distributive property.

Problem: $3x(2x+1)=0.$ Note, if you set 2x+1 equal to zero, and solve, you will get -1/2 this is also a solution.

You will distribute the number outside of the parenthesis to the numbers in the parenthesis. This means you will multiply 3x by 2x and by 1. so you get:

$6x^2+3x=0$

thus, x=0
x=-1/2

5. Originally Posted by bigcitydreams
In my Algebra class, we are learning about factoring. I really don't understand most of it and was wondering if someone could help and explain...
To learn the topic, try studying some online lessons on factoring quadratics.

Once you understand the general process, the worked solutions in your book and in this thread should make a lot more sense!

Have fun!