To find the product of (4x-5y)^2,
we can rewrite the problem as:
(4x-5y)(4x-5y) (two times because it is squared)
Now, time to use that old method we learned in middle school:
FOIL. (Firsts, Outers, Inners, and Lasts)
FOIL can help us greatly in this scenario.
Let's start by multiplying the 'Firsts' together:
4x * 4x = <em>16x^2</em>
Now, lets to the 'Outers':
4x * -5y = <em>-20xy</em>
Next, we can multiply the 'Inners':
-5y * 4x = <em>-20xy</em>
Finally, let's do the 'Lasts':
-5y * -5y = <em>25y</em>^2
Now, we can take the products of these equations from FOIL and combine like terms. We have: 16x^2, -20xy, -20xy, and 25y^2.
-20xy and -20xy make -40xy.
The final equation (product of (4x-5y)^2) is:
16x^2 - 40xy + 25y^2
Hope I helped! If any of my math is wrong, please report and let me know!
Have a good one.
Answer:
<h2>
<em>x</em><em>=</em><em>6</em></h2>
<em>Solution</em><em>,</em>
<em></em>
<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>
Critical values are values where f'(x)=0 and the bounds of a function. Thus, let's solve for f'(x)!
f(x)=2x^3-3x^2+3x+8
f'(x)=6x^2-6x+3
Now let's set f'(x)=0
0=6x^2-6x+3
0=2x^2-2x+1
As it turns out, 2x^2-2x+1 isn't factorable!
This saves me some time because this means there are no critical numbers!