# Homework Help: Factoring and solving for variable x

1. Feb 6, 2009

### imdapolak

1. Solve algebraically for...
1.) 4x*(x^2-5)+(x^2-5)^2=0 two separate problems not relating to each other 2.) 4x^3*(1-x)^3-3*(1-x)^2*x^4=0

2. this symbol ^ is to the power of. all x's are the x variable as written in my precalc book. For problem 1. I see that (x^2-5) is similar to (x^2-5)^2 but the ^2 outside of the brackets is throwing me off. Same goes for problem 2., i see that (1-x) is a common factor but the exponents are throwing me off. The variable x should have an answer using the zero product rule.

3.

Last edited: Feb 7, 2009
2. Feb 6, 2009

### NoMoreExams

For both of those, what factor is common in both parts i.e. for 1) if we let a = x^2 - 5 then we have 4x*a - a^2 = 0, can we factor anything out?

3. Feb 7, 2009

### epenguin

An apparent confusion of notation in your post, is one of the x's a multiply and is s the same as X?

4. Feb 7, 2009

### imdapolak

Fixed the problem, hope it clears any confusion. If not let me know.

5. Feb 7, 2009

### mplayer

When an exponent is outside of the parentheses, it is applied to the entire term inside of the parentheses.

In other words, (x2 - 5)2 = (x2 - 5)(x2 - 5)

6. Feb 7, 2009

### HallsofIvy

By the "distributive law" this is the same as (x^2- 5)(4x+ x^2- 5)= 0 so either x^2- 5= 0 or x^2+ 4x- 5= 0. x^2- 5= 0 has x= sqrt(5) and -sqrt(5) as roots. x^2+ 4x- 5= (x+5)(x-1)= 0 has x= 1 and x= -5 as roots.

Again, you can use the distributive law to say that is the same as (1-x)^2(4x^3(1-x)- 3x^4)= (1-x)^2(-4x^4+ 4x^3- 3x^4)= (1-x)^2(-7x^4+ 4x^3)= 0. Now it is easy to see that we can also factor out x^3: (1-x)^2(x^3)(-7x+ 4)= 0. We must have (1-x)^2= 0 or x^3= 0 or -7x+ 4= 0. The first of those has x= 1 as a double root. The second has x= 0 as a triple root and the last has x= 4/7 as root. Notice that counting "multiplicities" that gives us 6 roots. That is correct because if we were to multiply out the original polynomial, we would have a highest power term of x^6.

Last edited by a moderator: Feb 8, 2009