4(x-5)^2+2 i can not figure out what to do with this?

Answer 1

4(x-5)² + 2
4(x-5)(x-5) + 2
4(x² - 5x - 5x + 25) + 2
4(x² - 10x + 25) + 2
4x² - 40x + 100 + 2
4x² - 40x + 102

Answer 2

4(x-5)^2+2 = 4(x^2-10x+25)+2
= 4x^2 -40x+102
= 2(2x^2 -20x+51)


you don't have any context there. how's this though?

Answer 3

4(x-5)^2+2;
first off to factorise the equation ur using must intersect the x-axis, otherwise u use imaginary numbers, as this graph doesnt exist where y=0, you use complex (imaginary) numbers

first part - 4(x-5)^2 = (2)^2)(x-5)^2
= (2x-10)^2
now, Difference Of Perfect Squares tells us that
(a-b)(a+b) = a^2 - b^2

so, if we make a=(2x-10)
b=-2

then, (2x-10)^2 - ((√(-2))^2)
=(2x-10)^2 - (√2i)^2 i=√(-1)
=(2x-10-√2i)(2x-10+√2i)


im a bit rusty but i think that works