Respuesta :

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: [tex] x^4+9x^2-4x^2-36 [/tex]

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

Now you can rewrite this as [tex] (x^2-4)(x^2+9) [/tex] , however this is not completely factored. With (x^2 - 4), we are using the difference of squares, which is [tex] a^2-b^2=(a+b)(a-b) [/tex] . Applying that here, we have [tex] (x+2)(x-2)(x^2+9) [/tex] . 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: [tex] 2x^2+10x-x-5 [/tex]

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

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

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