# Thread: Issue with a factoring problem

1. ## Issue with a factoring problem

Factor completely.

all variables have an exponent which follows it

20x4 - 58x5 + 42x4

it would seem to me to factor a 2x to the fourth with a result of:

2x4(10x2 - 29x + 21)

again, all variables are followed by the exponent

Any help in understanding what the problem is would be of great appreciation. To me everything appears fine. If you distribute 2x to the fourth across, you end with the original expression.

However my homework web site tells me it's incorrect so I can only trust that they are right.

thanks again.

:Edited for clarity:

2. ## Re: Issue with a factoring problem

Originally Posted by helpwanted
Factor completely.

all variables have an exponent which follows it

20x4 - 58x5 + 42x4

it would seem to me to factor a 2x to the fourth with a result of:

2x4(10x2 - 29x + 21)

again, all variables are followed by the exponent

Any help in understanding what the problem is would be of great appreciation. To me everything appears fine. If you distribute 2x to the fourth across, you end with the original expression.

However this web site tells me it's incorrect so I can only trust that they are right.

thanks again.
You started with $\displaystyle 20x^4-58x^5+42x^4$

You then stated that this is equivalent to:

$\displaystyle 2x^4(10x^2 - 29x + 21)$

If we multiply out your expression, we get:

$\displaystyle 2x^4(10x^2)+2x^4(-29x)+2x^4(21)$

$\displaystyle =20x^6-58x^5+42x^4$

Which is not what you had originally.

Did you intend to write the problem statement as:

$\displaystyle 20x^6-58x^5+42x^4$?

Notice the quadratic in the brackets:

$\displaystyle 10x^2 - 29x + 21$
Taking the discriminant:

$\displaystyle 29^2-4(10)(21)$
$\displaystyle =1$

The discriminant is a perfect square, therefore we conclude that this quadratic, surprisingly, factors.

3. ## Re: Issue with a factoring problem

When multiplied out, one of the terms doesn't match.
Also, remember you can collect like terms before factoring.

4. ## Re: Issue with a factoring problem

Yes you are correct, I miss typed it, it is

20x^6-58x^5+42x^4

but I don't see the factors of:

2x^4(10x^2 - 29x + 21)

I'm now looking for factors of (10x^2 - 29x + 21).. and I'm failing to see the connection

5. ## Re: Issue with a factoring problem

Originally Posted by helpwanted
Yes you are correct, I miss typed it, it is

20x^6-58x^5+42x^4

but I don't see the factors of:

2x^4(10x^2 - 29x + 21)

I'm now looking for factors of (10x^2 - 29x + 21).. and I'm failing to see the connection
Factors of $\displaystyle 10x^2-29x+21$ are factors of $\displaystyle 2x^4(10x^2 - 29x + 21)$ and therefore, they are factors of $\displaystyle 20x^6-58x^5+42x^4$, which are what you have been asked to identify.

It's like saying, here's the number $\displaystyle 36$ and $\displaystyle 36=9\times{4}$ therefore the factors of $\displaystyle 36$ are $\displaystyle 9$ and $\displaystyle 4$. That's correct, and true, but it isn't complete. The factors of $\displaystyle 9$ and $\displaystyle 4$ are also factors of $\displaystyle 36$, amongst others.

For this quadratic, I'd suggest using the formula to find the roots, and deduce the factors from there, rather than use the "guess and check" method. I told you previously that the discriminant is 1, which should save you some time.

6. ## Re: Issue with a factoring problem

Well thank you for your time Quacky I really appreciate it. I edited my first post because it made it seem that mathhelpforum.com was telling me my answer was wrong, and what I meant to say was that my homework web site was, and aparently is correct at that. I however, do not see any way to further factor this problem more than I already have =/, I suppose that's why I'm always a C student in math.

-Park

7. ## Re: Issue with a factoring problem

I think it's more likely that you're a C student because you accept defeat too quickly!

$\displaystyle 10x^2-29x+21$

Take $\displaystyle f(x)=10x^2-29x+21$

Now then, when $\displaystyle f(x)=0$,

$\displaystyle 10x^2-29x+21=0$

In this situation, $\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}=0$

Now, you could go through and do the substitution - you'd end up with $\displaystyle x=\frac{29\pm{1}}{20}$

Either:

$\displaystyle x=\frac{30}{20}$ so $\displaystyle x=\frac{3}{2}$

or $\displaystyle x=\frac{28}{20}$ so $\displaystyle x=\frac{7}{5}$

Now, if a quadratic factors, then we have something of the form $\displaystyle (a'x+b')(cx+d)=0$ (the $\displaystyle '$ are just there to prevent repetition of $\displaystyle a$ and $\displaystyle b$)

Now we know that when $\displaystyle (a'x+b')(cx+d)=0$, $\displaystyle x=\frac{3}{2}$ firstly. This means that when $\displaystyle x=\frac{3}{2}$, one of those brackets is equal to $\displaystyle 0$.

I.e. $\displaystyle (a'\frac{3}{2}+b')=0$

You should soon be able to deduce that $\displaystyle a'=2$ and $\displaystyle b'=-3$ is one valid solution. Once you have that, working out (cx+d) should be even more straightforward because you know that $\displaystyle (2x-3)(cx+d)=10x^2-29x+21$ so by observation, $\displaystyle c=5$ and $\displaystyle d=-7$ to give us our factors:

$\displaystyle 10x^2-29x+21=(2x-3)(5x-7)$

Alternatively, you could use the guess-and-check method:
Factors of 10:
$\displaystyle 1,10$
$\displaystyle 2,5$

Factors of $\displaystyle 21$:
$\displaystyle 1,21$
$\displaystyle 3,7$

You need to take one pair of each numbers and multiply them together in such a way that the multiplied numbers:
-Add up/subtract to $\displaystyle 29$
-Multiply to $\displaystyle 21$

From there, you can just adjust the signs.

8. ## Re: Issue with a factoring problem

Ayee... I see now [light bulb!], thank for such a thorough response. The upper half of the response was above my current level, however the bottom portion makes perfect sense. For some reason when they said factor completely I was stuck with looking for the greatest common factor, and seeing none, I stopped, completely forgetting to factor the trinomial.

Your response and time is very much appreciated.

9. ## Re: Issue with a factoring problem

Originally Posted by helpwanted
Ayee... I see now [light bulb!], thank for such a thorough response. The upper half of the response was above my current level, however the bottom portion makes perfect sense. For some reason when they said factor completely I was stuck with looking for the greatest common factor, and seeing none, I stopped, completely forgetting to factor the trinomial.

Your response and time is very much appreciated.
That's a fairly common trail of thought; don't be disheartened by it. In fact, it's a very deceiving question: $\displaystyle 10x^2-29x+21$ isn't something that looks as though it factors - I wouldn't have spotted the factors if not for the fact that your homework checker caught it out.

10. ## Re: Issue with a factoring problem

Ok; one a little easier (so you feel better!); factor completely:
5x^6 - 25x^5 + 30x^4

11. ## Re: Issue with a factoring problem

I have another factoring issue. 32 has an exponent of 2.

[PHP].08r - .04 = -.32r^2[/PHP]

multiply by 100 to get rid of the decimals:

[PHP]8r - 4 = 32r^2[/PHP]

set the equation equal to zero:

[PHP]32r^2 + 8r - 4 = 0[/PHP]

factor the greatest common factor:

[PHP]
4(8r^2 + 2 - 1) = 0

//--->or

-4(-8r^2 - 2 +1) = 0[/PHP]

The above does not factor for me no matter how I try =/. The question is expecting two answers for R. A smaller number, and a Larger number. Thanks again.

12. ## Re: Issue with a factoring problem

Hello, helpwanted!

$\displaystyle \text{Factor completely: }\:20x^6 - 58x^5 + 42x^4$

Factor out $\displaystyle 2x^4\!:\quad 2x^4(10x^2 - 29x + 21)$

Do you mean you have never been taught how to factor a trinomial?

Here's a method you might like . . . or maybe not.

Multiply the leading coefficient (10) by the constant (21): $\displaystyle (10)(21) \,=\,210$

Factor 210 into two parts whose sum is the middle coefficient (29).

How do we find that pair?
Divide by 210 by 1, 2, 3, etc., keeping the ones that "come out even",
. . and look for the pair whose sum is 29.

. . $\displaystyle \begin{array}{cccc}\text{Divide} & \text{Factors} & \text{Sum} \\ \hline 210 \div 1 & 1\cdot210 & 211 \\ 210 \div 2 & 2\cdot105 & 107 \\ 210 \div 3 & 3\cdot70 & 73 \\ 210 \div 4 & --- & -- \\ 210 \div 5 & 5\cdot42 & 47 \\ 210 \div 6 & 6\cdot35 & 41 \\ 210 \div 7 & 7\cdot30 & 37 \\ 210 \div 8 & --- & -- \\ 210 \div 9 & --- & -- \\ 210 \div 10 & 10\cdot21 & 31 \\ 210 \div 11 & --- & -- \\ 210 \div 12 & --- & -- \\ 210 \div 13 & --- & -- \\ 210 \div 14 & 14\cdot15 & 29 \\ && \uparrow \\ && \text{There!} \end{array}$

Replace the middle term $\displaystyle (-29x)$ with $\displaystyle -14x$ and $\displaystyle -15x.$

We have: .$\displaystyle 10x^2 - 14x - 15x + 21$

Factor by grouping:
. . . . . . . $\displaystyle 2x(5x - 7) - 3(5x - 7)$

Factor: .$\displaystyle (5x-7)(2x-3)$

13. ## Re: Issue with a factoring problem

Thanks Soroban. I find my mind likes to work mathmatical problems out through procedure, and for some reason while factoring I have a really hard time maintaining a procedure that brings me to the correct outcome. I will give your example a shot in a few problems.

Thanks again.

14. ## Re: Issue with a factoring problem

Originally Posted by helpwanted
4(8r^2 + 2 - 1) = 0
That should be 8r^2 + 2r - 1 = 0

Use the quadratic formula to solve:
r = [-2 +- SQRT(2^2 - 4(8)(-1)] / [2(8)]

You seem unaware of the basics; attending math classes?