How do you factor completely #x^4 + 7x^2 + 10#?

1 Answer
Jan 3, 2016

Answer:

You need to change the variable in order to find the roots.

Explanation:

Let's say #P(x) = x^4 + 7x^2 + 10# and let's say #X = x^2#. So we're now studying #P(X) = X^2 + 7X + 10#, which is a classic trinomial we can solve easily.

In order to use the quadratic formula, we first need to calculate #Delta = b^2 - 4ac#. Here, #Delta = 7^2 -4*10 = 9#. So this polynomial has 2 real roots.

By the quadratic formula, the roots are given by #(-b +- sqrtDelta)/2a#.

#X_1 = (-7 - 3)/2 = -10/2 = -5# and #X_2 = (-7 + 3)/2 = -4/2 = -2#.

So #P(X) = (X+5)(X+2)#. We now switch back to the original variable, hence #P(x) = (x^2 + 5)(x^2 + 2)#. If we want to factorise it completely, there is still some work to do with complex numbers.

I will only show you how to solve #x^2 + 5 = 0# in #CC# because it's the exact same way to solve #x^2 + 2 = 0#.

In #CC#, #x^2 + 5 = 0 iff x^2 = -5 iff x = +-isqrt5#. So the complete factorization of #P# in #CC# is #(x - isqrt5)(x + isqrt5)(x - isqrt2)(x + isqrt2)#.