help on algebra 2 homework?

Question

help with algebra 2 hw?

can someone provide a step by step explanation on how to solve the following problems please because i dont understand:

4x^4+39x^2-10

8x^3-64

3x^4-24x

3x^4+9x^3+x^2+3x

also, u must factor to get the answers so (i looked up the answer to this problem):

(2x^2+3)(9x-1) would be the answer to:

18x^3-2x^2+27x-3

if anyone could help me by showing me how to do one or all of these problems id be very grateful

Answer 1

OK...the first one is a lot like this problem--

4y^2 +39y -10
this factors into (4y-1) (y+10)

the problem you had for #1 is JUST like this, but everywhere *I* had a "y", you had an x^2.

so....

4x^4 + 39x^2 -10

factors to (4x^2 -1)(x^2 + 10)

the first term is the difference of 2 squares.... so

(2x+1)(2x-1)(x^2+10) because x^2-a^2=(x+a)(x-a) for the difference of 2 squares


for #2... 8x^3 - 64... the difference of 2 cubes

or factor out an 8 and still the difference of 2 cubes 8(x^3 - 8)

8(x - 2)(x^2 + 2x +4) the formulae for diff of 2 squares and diff of 2 cubes should be in your text



3. 3x^4 - 24x factor out 3x....

3x (x^3 - 8)

again, the difference of 2 cubes

3x(x - 2)(x^2 + 2x + 4)

4. 3x^4 + 9x^3 + x^2 +3x

factor out an x

x(3x^3 + 9x^2 + x + 3)

now factor the 2nd term... you can see the first two parts have to be 3x^2 and x.

find the others to multiply out to 3 and give you the 9x^2 term and the x term...

x(3x^2 + 1)(x + 3)


as you know, for all 4 problems you set what you factored = 0 and see what values for x would give 0 in each term.

for that last one,
x could be 0 all by itself
3x^2 + 1 could be 0, in which case x is imaginary sqrt of -1/3

or x could = -3, making the 3rd term = 0


hope all that helps!!!

Answer 2

I'll show you #2. That should allow you to solve the others:

factor out an 8: 8 (x^3 - 8). The binomial is the difference of perfect cubes, which is factored as (x-2)(x^2 + 2x + 4)

the answer becomes 8 (x-2) (x^2 +2x + 4)

Answer 3

Well I am in Algebra II also, I have seen something a little close to what you mentioned. For my class we would change the problem into logrithmic form. I dont know if you have covered that or not.