Factoring Problem?

Question

Factor the polrnomial

4x^4 - 5x^2 - 9

***Please show all work!***

Answer 1

Let u = x^2. Factor:
4u^2 - 5u - 9 = (4u - 9)(u + 1)
= (4x^2 - 9)(x^2 + 1) [Substitute x^2 for u]
= (2x - 3)(2x + 3)(x^2 + 1) [Difference of Squares]

Answer 2

Substitute 1 for x:

4 - 5 - 9 = -10.

No that doesn't work. Try -1:

4 + 5 - 9 = -10. No work. We need to change the sign of the middle term. Only an imaginary number can do it.

Try i, the square root of -1:

4i^4 - 5i^2 - 9 = 4+5-9 = 0

That does work. Therefore, -i is also a root, since complex roots come in conjugate pairs. Multiply this:

(x - i) (x - (-i)) = (x - i) (x + i) = x^2 + 1.

So x^2 + 1 is a factor. Divide it into 4x^4 - 5x^2 - 9:

4x^2 - 9
x^2 + 1 | 4x^4 - 5x^2 - 9
4x^4 + 4x^2
_________
-9 x^2 - 9
-9x^2 - 9
_________
0

So 4x^2 - 9 is a factor also. This is a difference of two squares: (2x + 3)(2x-3). Put it all together:

4x^4 - 5x^2 - 9
= (x^2 + 1) (4x^2-9)
=(x^2 + 1) (2x+3)(2x-3) over the real numbers
= (x+i)(x-i)(2x+3)(2x-3) over the complex numbers

Answer 3

4x^4 - 5x^2 - 9
lets do something simpler by sub'g y = x^2
so a simpler form looks like this
4y^2 -5y -9
so factoring this looks like (1y +1)(4y-9)
now replace y with x^2 and you get
(x^2+1)(4x^2-9)
you should also see that (4x^2-9) = (2x+3)(2x-3)
so now (x^2+1)(2x+3)(2x-3)
x= i or sqrt of -1 {from the x^2 +1}
x= 3/2 and -3/2

Summary x= i, 3/2, -3/2

Answer 4

4x^4 - 5x^2 - 9 = 0
(4x^2-9)(x^2+1) = 0
4x^2-9 = 0 or x^2+1 = 0
4x^2 = 9 or x^2 = -1
x^2 = 9/4 or x^2 = -1
x = rootsquare(9/4) or x = rootsquare(-1)---> impossible
so x = rootsquare(9/4)
x = 3/2 or -3/2

solved!!!

Answer 5

(x-9)(x+4) because -9 times 4 equals 36 so you put that at the top of the diamond problem and then -5 at the bottom, and since -9 +4 equals -5 when added and -36 when multiplied, this has to be correct. (hopefully)

Answer 6

4x^4 - 5x^2 - 9
(4x-9)(x+1) that's all the work to show.

Answer 7

Fortman has the right answer except he should not say the
sqrt(-1) is impossible. He should have said theat the other two roots are i and -i. A 4th degree polynomial must have four roots, even if two of them or all of them are imaginary.

Answer 8

4x^4 - 5x^2 - 9 (start)
256x - 25x - 9 (apply exponents)
231x - 9 (subtract x's) (answer)