How do you factor x^4 -10x^2 +9?

Question

I totally forgot, all I know is that the answer should look like:

f(x) = (_____)(_____)(_____)(_____)

Answer 1

A recap on Remainder and Factor theorem:
If f(a) = 0, then (x-a) is a factor of f(x).

We know that the term in f(x) independent of x is a product of all the roots of f(x), so we can investigate the factors of 9 [+-1, +-3, +-9] in this case to find out the 1st factor of f(x).

By trial and error, f(3) = 81-90+9 = 0, so (x-3) is a factor of f(x)
Using long division f(x) is factorised as follows:
f(x) = (x - 3)(x^3 + 3x^2 - x - 3)

So the problem reduces to finding factors for a g(x) = x^3 + 3x^2 - x - 3.

Thus similarly, by trial and error, f(1) = 1+3-1-3=0, so (x-1) is a factor of g(x) and therefore f(x) as well.

So our f(x) becomes f(x) = (x - 3)(x - 1)(x^2 + 4x + 3),
where x^2 + 4x + 3is quadratic and so can be easily factorised to (x + 3)(x + 1).

Therefore, f(x) = (x - 3)(x - 1)(x + 3)(x + 1)
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Alternatively, if you observe that f(x) consists of only even powers of x, you may try to use substitution method to factorise f(x).

Let y = x^2,

therefore f(x) = x^4 -10x^2 +9 becomes f(y) = y^2 - 10y + 9, which is quadratic and can be easily factorised to give

f(y) = (y - 9)(y - 1)

Substituting x back, we get

f(x) = (x^2 - 9)(x^2 - 1) = (x + 3)(x - 3)(x + 1)(x - 1),

using the property a^2 - b^2 = (a + b)(a - b).

Answer 2

x^4 - 10x^2 + 9
(x^2 - 1)(x^2 - 9)
(x - 1)(x + 1)(x - 3)(x + 3)

you can just treat the problem like x^2 - 10x + 9 at first

Answer 3

Write y for x^2, then factor
y^2 - 10y + 9

Then, in the two factors, write x^2 in place of y, and each factor will be a difference of squares, which you can factorise into two factors each.

Answer 4

f(x)=x^4 -10x^2 +9
f(x)=(x^2-9)(x^2-1)
f(x)=(x+3)(x-3)(x+1)(x-1)