Question

Factor both quadratic expressions. (x 4 + 5x 2 - 36)(2x 2 + 9x - 5) = 0

Answer 1

So focusing on x^4 + 5x^2 - 36, we will be completing the square. Firstly, what two terms have a product of -36x^4 and a sum of 5x^2? That would be 9x^2 and -4x^2. Replace 5x^2 with 9x^2 - 4x^2:
`x^4+9x^2-4x^2-36`

Next, factor x^4 + 9x^2 and -4x^2 - 36 separately. Make sure that they have the same quantity inside of the parentheses:
`x^2(x^2+9)-4(x^2+9)`

Now you can rewrite this as
`(x^2-4)(x^2+9)` , however this is not completely factored. With (x^2 - 4), we are using the difference of squares, which is
`a^2-b^2=(a+b)(a-b)` . Applying that here, we have
`(x+2)(x-2)(x^2+9)` . x^4 + 5x^2 - 36 is completely factored.

Next, focusing now on 2x^2 + 9x - 5, we will also be completing the square. What two terms have a product of -10x^2 and a sum of 9x? That would be 10x and -x. Replace 9x with 10x - x:
`2x^2+10x-x-5`

Next, factor 2x^2 + 10x and -x - 5 separately. Make sure that they have the same quantity on the inside:
`2x(x+5)-1(x+5)`

Now you can rewrite the equation as
`(2x-1)(x+5)` . 2x^2 + 9x - 5 is completely factored.

Putting it all together, your factored expression is
`(x+2)(x-2)(x^2+9)(2x-1)(x+5)=0`