Factoring and solving for variable x

[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.

 

[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?

 

[epenguin]

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

 

[imdapolak]

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

 

[mplayer]

For problem 1. I see that (x²-5) is similar to (x²-5)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.

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)

 

[HallsofIvy]

1. Solve algebraically for...
1.) 4x*(x^2-5)+(x^2-5)^2=0

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.

2.) 4x^3*(1-x)^3-3*(1-x)^2*x^4=0

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.

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.