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](https://tex.z-dn.net/?f=%20x%5E4%2B9x%5E2-4x%5E2-36%20)
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)](https://tex.z-dn.net/?f=%20x%5E2%28x%5E2%2B9%29-4%28x%5E2%2B9%29%20)
Now you can rewrite this as
, however this is not completely factored. With (x^2 - 4), we are using the difference of squares, which is
. Applying that here, we have
. 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](https://tex.z-dn.net/?f=%202x%5E2%2B10x-x-5%20)
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)](https://tex.z-dn.net/?f=%202x%28x%2B5%29-1%28x%2B5%29%20)
Now you can rewrite the equation as
. 2x^2 + 9x - 5 is completely factored.
<h3><u>Putting it all together, your factored expression is
![(x+2)(x-2)(x^2+9)(2x-1)(x+5)=0](https://tex.z-dn.net/?f=%20%28x%2B2%29%28x-2%29%28x%5E2%2B9%29%282x-1%29%28x%2B5%29%3D0%20)
</u></h3>
Answer:
x = 28
Step-by-step explanation:
(3x + 6^o) = 90^o
3x = 84^o
x = 28
![\quad a^2-10a+16-6b-b^2](https://tex.z-dn.net/?f=%20%5Cquad%20a%5E2-10a%2B16-6b-b%5E2)
$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$
Answer:
zero slope
Step-by-step explanation: