How do you factor x^4-8x^2+15?

Question

how do you factor x^4-8x^2+15 and -6q^5-6000q^2?

Answer 1

x^4 - 8x^2 + 15
assume a = x^2. now you have
a^2 - 8a + 15 = (a - 3)(a - 5)

substituting x^2 back in:
(x^2 - 3)(x^2 - 5)

neither of those qualify for the difference of squares formula, so you're pretty much done factoring.
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-6q^5 - 6000q^2

each term has a (-6) and q^2, so factor those out:

(-6q^2)(q^3 + 1000)

Note: 1000 = 10^3, so you can use the sum of cubes formula to factor (q^3 + 1000)

(-6q^2)(q + 10)(q^2 - 10q + 100)

Answer 2

Substitute x^2 by say t. You then have a quadratic equation in t which tou can solve by the usual method. Then each t = x^2 will give rise to 2 values for x and you'll have 4 roots for the 4th degree equation, as it should be in the general case.

Answer 3

Question 1
= (x² - 5).(x² - 3)
= (x - √5).(x + √5).(x - √3).(x + √3)

Question 2
- 6.q² (q³ + 1000)

Answer 4

(x^2-3)(x^2-5)

-6q^2(q^3+1000)

Answer 5

you search the roots of equation given x^2 = X

so X^2 -8X +15 =0 you have X =5 and X=3

this yields for x1 = sqrt5 x2 = -sqrt5 x3 = sqrt3 x4 =-sqrt3

and the result is (x-sqrt5)(x+sqrt5)(x-sqrt3)(x+sqrt3)

you put -2q^2 in factor

-2q^2 (3q^3+2000)

Answer 6

1) x^2(x^2-8)

2) -6q^2(q^3+1000)

Answer 7

a million. 3(x-8)(x+8) 2. 2(9x+2)(9x-2) 3.3(x-2)(x-12) 4. 3(x-7)(x-3) 5.(4x-a million)(4x-a million) 6.3(x-12)(x-2) 7.4(x-8)(x+10) 8. 2(x+2)(x+3) 9. 3(x+2)(x-a million) 10. (x+a million)(x-a million)(2x+3) 11.3(x+a million)(x+3)(x-3) 12. (x-2)(x+2)(x^2+4) 13. (y-3)(y+3)(y^2+9) 14. 2(2x+3)(2x-3) 15. 2(x-4)(x-4)