How do I Factor This Equation?

Question

Please help me factor this equation completely, please show steps.

x^4 - 10x^2 + 9 = 0

Thank you!

Answer 1

place x^2 = y => x^4 = y^2

y^2- 10y + 9 = 0 => y= 1 or y= 9

y=1 => x = +-1

y=9 => x= +- 3

Answer 2

Are u fifth grader? I solved these types when I was in Fifth, but then Maths is very hard taught in india..

yup all the answers are right..

Answer 3

(x^2 - 9)(x^2 - 1) = 0

(x-3)(x+3)(x-1)(x+1)=0

x=3,-3,1,-1

Answer 4

(x^2 -9)(x^2 -1)=0

by the zero property theorum

either(x^2 -9) =0 or (x^2 -1)=0

(x^2 -9) =0

(x-3)(x+3)=0 so x=+3 or -3

(x^2 -1)=0

(x-1)(x+1)=0

so x=+1 or x=-1

Answer 5

(x^2-9)(x^2 -1)=0

(x-3)(x+3)(x-1)(x+1)=0

Answer 6

x^4 - 10x^2 + 9 = 0

(x^2 - 9)(x^2 - 1) = 0

(x^2 - 3^2)(x^2 - 1^2) = 0

(x - 3)(x + 3)(x - 1)(x + 1) = 0

Answer 7

If you let y = x^2, the equation becomes

y^2 - 10y + 9 = 0

which factors as

(y - 1)(y - 9) = 0

Replacing y, we have

(x^2 - 1)(x^2 - 9) = 0

Both of these are the difference of two squares

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

(x^2 - 9) = (x - 9)(x + 9)

So the complete factorization is

(x+1)(x - 1)(x - 9)(x + 9) = 0

Answer 8

x^4 - 10x^2 + 9 = 0

x^4 - x^2 - 9x^2 + 9 = 0

(x^4 - x^2) - (9x^2 - 9) = 0

x^2(x^2 - 1) - 9(x^2 - 1) = 0

(x^2 - 1)(x^2 - 9) = 0

(x + 1)(x - 1)(x + 3)(x - 3) = 0

x + 1 = 0

x = -1

x - 1 = 0

x = 1

x + 3 = 0

x = -3

x - 3 = 0

x = 3

∴ x = ±1 , ±3