Factoring by Using a Quadratic Form
Problem: x^4 + 8x^2 - 20
What I have so far:
(x^2)^2 + 8(x^2) - 20
a^2 + 8a - 20
For some reason my textbook says to substitute the (x^2) with (a) as you can see in the las step. The next step after that is factoring [a^2 + 8a - 20].
x^4 + 8x^2 - 20
(x^2 - 2)(x^2 + 10)
To factor the expression x^4 + 8x^2 - 20 using the quadratic form, you need to substitute (x^2) with a new variable, let's say 'a'. This substitution allows you to rewrite the expression in a quadratic form, which can be factored more easily.
So, by substituting (x^2) with (a), your original expression becomes a^2 + 8a - 20. Now, we can proceed with factoring a^2 + 8a - 20.
To factor the quadratic expression a^2 + 8a - 20, you need to find two numbers, let's call them 'm' and 'n', such that m and n satisfy the equation m * n = a * (-20), and m + n = 8a.
In other words, you are looking for two numbers whose product is -20, and their sum is 8. By applying trial and error or any numerical method, you can find that the numbers 10 and -2 satisfy these conditions: 10 * (-2) = -20, and 10 + (-2) = 8.
Now, you can rewrite the expression a^2 + 8a - 20 as follows:
a^2 + 10a - 2a - 20
Next, you group the terms together:
(a^2 + 10a) + (-2a - 20)
Then, factor out the greatest common factor from each group:
a(a + 10) - 2(a + 10)
Now, notice that we have a common binomial factor, (a + 10), which can be factored out:
(a + 10)(a - 2)
Finally, substitute back the original variable (x^2) in place of 'a':
(x^2 + 10)(x^2 - 2)
Therefore, the factored form of the expression x^4 + 8x^2 - 20 is (x^2 + 10)(x^2 - 2).