Someone smart who knows anything about algebra, please help!?

Question

x^5-x^3... and how would I go about factoring it?

x^3 (x^2)... that's what I would have done, or rather, did do... that's wrong, isn't it?

ALSO, what would a problem like this:

Solve for x:
(SQRT 18x-5) = x+4

what would that be called? I have a book that can explain to me every type of problem, I just don't know what to look it up under, thanks for all the help, I really appreciate it!

Answer 1

x^3 (x^2 - 1 )

Don't forget that when you factored out x^3, 1s left behind!

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(SQRT 18x-5) = x+4

Square both sides.
18x - 5 =(x +4) ^2
18x - 5 = x^2 + 8x +16

Move everything to one side of the equals, leaving zero.
0 = x^2 -10x +21

Factor
0 = (x-7) (x - 3)

Answers then are the opposites of the numbers above...

x = +3, +7

Answer 2

JG's answers are correct but incomplete.

x^5 - x^3 = x^3(x^2 - 1)

There are patterns you need to learn to recognize. One of them is the difference of squares, which is what x^2 - 1 is. The complete factorization would be

x^5 - x^3 = x^3(x + 1)(x - 1)

The difference of squares pattern is

x^2 - a^2 = (x + a)(x - a)


For the other - yes, square both sides. However, since SQRT() is only positive but squaring a negative gives you the same result, you have a chance of introducing extraneous roots into an equation when you square both sides. Extraneous roots are values that work when the equation is squared but don't work in the original.

sqrt(18x - 5) = x+4

18x - 5 = x^2 + 8x + 16

x^2 -10x + 21 = 0

(x - 3)(x - 7) = 0

x = 3 or 7

To check for extraneous roots, plug the answers back into the original equation and check that they both work.

sqrt(18*3 - 5) = 3 + 4

sqrt(54 - 5) = 7

sqrt(49) = 7 and 3 works.

sqrt(18*7 - 5) = 7 + 4

sqrt(126 - 5) = 11

sqrt(121) = 11 and 7 works as well.

So, in this case, both roots work, but it's a really bad idea to assume that they always will.

Answer 3

The previous person answered the second question.

They left out a bit more for the first question:
x^5 - x^3 = x^3 (x^2 - 1) = x^3 (x -1) (x + 1)

Answer 4

x^5 - x^3 = x^3(x^2 - 1) = x^3(x - 1)(x + 1)

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sqrt(18x - 5) = x + 4
18x - 5 = (x + 4)^2
18x - 5 = (x + 4)(x + 4)
18x - 5 = x^2 + 4x + 4x + 16
18x - 5 = x^2 + 8x + 16
x^2 - 10x + 21 = 0
(x - 7)(x - 3) = 0
x = 7 or 3

Answer 5

For the first one, think of the distributive property: ab + ac = a (b + c) or, in your case: ab - ac = a (b - c) and then use it to factor out your common factors. Many people have already solved it for you. Also, as one person already pointed out, you need to look inside of the parenthesis to see if anything else can be factored out, which in your case was the difference of two squares.

Sorry, I don't remember what the second equation is called.

Answer 6

x^5-x^3=x^39x^2-1)=x^3(x+1)(x-1)

sqrt(18x-5)=x+4
square both sides
18x-5=(x+4)^2=x^2+8x+16
x^2+8x-18x+16+5
x^2-10x+10
(x-3)(x-7)

x=3
or
x=7

check
sqrt(18*7-5)=7+4
sqrt(216-5)=11
sqrt(121)=11
11=11

sqrt(18*3-5)=3+4
sqrt(54-5)=7
sqrt(49)=7
7=7

Answer 7

same answer as before:
http://answers.yahoo.com/question/index;_ylt=AjldaYdUwBd_JEuW85dkCOPzy6IX?qid=20060928183353AAxuiGi