Factoring Polynomials? Maybe. More like word problems but take a look.

• Nov 26th 2012, 08:46 AM
Elton20
Factoring Polynomials? Maybe. More like word problems but take a look.
It says, A positive number is 30 less than its square. Find the number.
Anybody have a clue on how to write the equation?
Find two consecutive negative integers whose product is 90
The sum of the squares of two consecutive positive even integers is 340. Find the integers.
Those three are not my specialty so I just need a little clarification. Thanks.
• Nov 26th 2012, 08:56 AM
Re: Factoring Polynomials? Maybe. More like word problems but take a look.
A positive number is 30 less than its square. Find the number.

First, let us label the unknown, the positive number we are asked to solve. For example, we can use "n". Hence n^2 is its square. What does it mean to be 30 less? Another way to phrase is that it means that the square, n^2, exceeds n by 30. So if we add 30 to n we get n^2. In equation form this can be written as

n+30 = n^2
n^2 - n - 30 = 0

Here you should get two solutions. Since the problem says 'positive number', this will limit your solution.
--

For the other two, I'll give you something to start with and you can write the equations on your own.

2) Consecutive can colloquially mean "the term in the sequence that follows". So if we choose n again as our integer, n+1 is its consecutive number. Products between two numbers can be written as xy, where x and y are two numbers.

3) Trickier in terms of wording, but same concepts apply as above.

What does it mean to have a sum of squares? ex. x^2 + y^2
What does it mean to have two consecutive positive even integers? ex. 2 and 4 and 2n and 2(n+1) [can you justify the last example?]

If you want me to verify if you set up the equations correctly, you may ask.
• Nov 26th 2012, 09:09 AM
Elton20
Re: Factoring Polynomials? Maybe. More like word problems but take a look.
well how do I find the number for the first problem.when I solve for n, how do I do it?
• Nov 26th 2012, 09:12 AM
Elton20
Re: Factoring Polynomials? Maybe. More like word problems but take a look.
also for second equation, I wrote as n^2 - n=90. Do I solve further of what? I do not know how to solve with squares unless its square rooting both sides which gives me a wrong answer according to the back of the book I'm using.
• Nov 26th 2012, 09:25 AM
Re: Factoring Polynomials? Maybe. More like word problems but take a look.
Alright, here's the case of factoring a quadratic polynomial when the coefficient of $\displaystyle n^2$ is 1.
The goal here is to express the polynomial as the product of two terms, namely $\displaystyle (n+a)(n+b)$
If we can do this then the equation becomes $\displaystyle (n+a)(n+b) = 0$

What happens if $\displaystyle n = -a$?

$\displaystyle (n+a)(n+b) = (-a+a)(-a+b) = 0(-a+b)$ and since 0 times anything is 0, then $\displaystyle 0(-a+b) = 0$

Similar case for $\displaystyle n = -b$

Hence $\displaystyle n= -a, -b$ are the solutions to $\displaystyle (n+a)(n+b) = 0$ and hence are the solutions to, when we foil, $\displaystyle (n+a)(n+b) = n^2 + (a+b)n + ab = 0$

Now let us consider $\displaystyle n^2-n-30$
Our goal is to make it look like $\displaystyle n^2 + (a+b)n + ab$

What must a and b be?
By comparison, we have that $\displaystyle -1 = a+b$ [these are the coefficients of n] and $\displaystyle -30 = ab$ [this is the constant in both polynomials.]
You can solve this system using algebra or inspection. By inspection, we get a=5 b=-6

Hence $\displaystyle n^2-n-30 = n^2 + (5-6)n + 5(-6) = (n+5)(n-6) = 0$
Hence $\displaystyle n= -5, 6$

Since we are limited to only positive solutions, $\displaystyle n=6$
• Nov 26th 2012, 09:29 AM
Plato
Re: Factoring Polynomials? Maybe. More like word problems but take a look.
Quote:

Originally Posted by Elton20
also for second equation, I wrote as n^2 - n=90. Do I solve further of what? I do not know how to solve with squares unless its square rooting both sides which gives me a wrong answer according to the back of the book I'm using.

Have you mastered middle school algebra?
If not, then start there and learn the basics.

If you have then you should be able to do these simple examples.
• Nov 26th 2012, 09:32 AM
Re: Factoring Polynomials? Maybe. More like word problems but take a look.
Quote:

Originally Posted by Elton20
also for second equation, I wrote as n^2 - n=90. Do I solve further of what? I do not know how to solve with squares unless its square rooting both sides which gives me a wrong answer according to the back of the book I'm using.

Tell me your thought process behind writing $\displaystyle n^2-n=90$
• Nov 26th 2012, 09:48 AM
Elton20
Re: Factoring Polynomials? Maybe. More like word problems but take a look.
nope, not mastered middle school algebra. I'm in it, thats the reason I'm asking. so the problem was Find two consecutive negative integers whose product is 90, the equation was written as n(n-1)=90. Isnt this right? So I simplified into n^2-n=90. I put it as n(n-1) because n is one integer the consecutive negative should be negative one since it's consecutive negative. There you go.
• Nov 26th 2012, 09:55 AM
Plato
Re: Factoring Polynomials? Maybe. More like word problems but take a look.
Quote:

Originally Posted by Elton20
nope, not mastered middle school algebra. I'm in it, thats the reason I'm asking. so the problem was Find two consecutive negative integers whose product is 90, the equation was written as n(n-1)=90. Isnt this right? So I simplified into n^2-n=90. I put it as n(n-1) because n is one integer the consecutive negative should be negative one since it's consecutive negative. There you go.

As a general principle, these sorts of questions are dealt with the zero property.
If $\displaystyle a\cdot b=0$ then $\displaystyle a=0\text{ or }b=0~.$

To use that property on $\displaystyle n(n-1)=90$ we need $\displaystyle n^2-n-90=0$.

Now use your factoring ability to factor. Then use zero property.
• Nov 26th 2012, 09:59 AM
Elton20
Re: Factoring Polynomials? Maybe. More like word problems but take a look.
Ah, I see. Ok thank you.