Question

How do you solve `8x^5 + 10x^4 = 4x^3 + 5x^2`?

Answer 1

`x=0 or x=-5/4 or x=+-sqrt(1/2)`

Explanation:

Move all the terms to the left.

`8x^5+10x^4-4x^3-5x^2=0`

Group in pairs and factor each pair.

`(8x^5+10x^4)+(-4x^3-5x^2)=0`

`2x^4(4x+5)-x^2(4x+5)=0`

Factor out the common factor and common bracket:

`x^2(4x+5)(2x^2-1)=0`

There are now three factors. Set each equal to zero and solve.

`x^2 = 0" "rarr x =0`

`4x+5=o" "rarr x =-5/4`

`2x^2-1=0" "rarr x^2 = 1/2" "rarr x = +-sqrt(1/2)`

Answer 2

`x = 0, pm 1/sqrt{2}, -5/4`

Explanation:

We can easily rewrite `8x^5 + 10x^4 = 4x^3 + 5x^2` in the form

`2x^4(4x+5) = x^2(4x +5)`

Subtracting `x^2(4x +5)` from both sides, leads to

`2x^4(4x+5) - x^2(4x +5) = 0`

which means that
'
`(2x^4 -x^2)(4x+5)=0`,

i.e.

`x^2(2x^2-1)(4x+5)=0`

Now, for a polynomial to vanish, one of its factors have to be zero. So, either `x^2=0`, or `2x^2-1=0` or `4x+5=0`.

If `x^2=0`, we must have `x=0`

or if `2x^2-1=0`, we get `x^2=1/2`, i.e. `x=pm 1/sqrt{2}`

or, finally, `4x+5=0` which implies that `x=-5/4`