An expression containing the fourth power of x, that is x^4, can be factored just like a quadratic expression. For example:
`x^4 + 3x^2 + 2`
The above expression contains x^4, but it can be considered a quadratic expression and factored just like quadratic expressions. For a moment, forget that there is an x^4 in it. Just look at the the numbers in the expression. Comparing with standard quadratic expression `ax^2 + bx + c`,
`x^4 + 3x^2 + 2`
The above expression contains x^4, but it can be considered a quadratic expression and factored just like quadratic expressions. For a moment, forget that there is an x^4 in it. Just look at the the numbers in the expression. Comparing with standard quadratic expression `ax^2 + bx + c`,
- `a = 1`
- `b = 3`
- `c = 2`
Product of `a` and `c` is 2.
So you need to find two numbers whose product is 2 and sum is 3. Try to find them first without looking at the solution below. Let me leave some blank space between this and the solution so that you don't see the answer before trying to solve it :~)
If you were able to solve it, good! The two numbers are 1 and 2, because 1 times 2 is 2 and 1 plus 2 is 3. Now replace `3x` in the equation (which is the middle term) with these numbers `2x + x` to get
`x^4 + 2x^2 + x^2 + 2`
Rearrange the terms so that the two two's are together,
`x^4 + x^2 + 2x^2 + 2`
Factor out `x^2` from the first two terms and 2 from the next two terms
`x^2(x^2 + 1) + 2(x^2 + 1)`
Factor out `(x^2 + 1)` from the expression,
`(x^2 + 1)(x^2 + 2x)`
~ * ~
Thus we factored an expression having a degree of four (that is x^4) as if it was a quadratic expression because it had
- three terms in it
- did not contain an x^3
So we notice that we can factor all three termed expressions, called trinomials, as long as the degree is even, there are three terms, the second term has a degree half that of the first term, and the third term is either a constant or has a degree that is half that of the second term. So if you have x^6 + 3x^3 + 2, you can factor it just like you factored the x^4 one above, as follows:
`x^6 + 3x^3 + 2 = x^6 + x^3 + 2x^3 + 2 = x^3(x^3 + 1) + 2(x^3 + 1) = (x^3 + 1)(x^3 + 2)`
Similarly if you have an expression with `x^8` you can factor it out as follows:
`x^8 + 3x^4 + 2 = x^8 + x^4 + 2x^4 + 2 = x^4(x^4 + 1) + 2(x^4 + 1) = (x^4 + 1)(x^4 + 2)`
Summary
The above method is called splitting the middle term. You can factor any trinomial by the method of splitting the middle term as long as its highest power (degree) is even, degree of its second term is half that of the first one and the third term is either a constant or has a degree that is half of the second term's degree.
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