What is the standard form of # y= (4x-15)(2x-2)-(3x-1)^2#?

1 Answer
Apr 7, 2018

#y = -x^2 - 32x + 29#

Here's how I did it:

Explanation:

Standard form means that we have to put the equation in this form: #y = ax^2 + bx + c#.

#y = (4x-15)(2x-2)-(3x-1)^2#

The first thing we have to do is distribute and expand:
#4x * 2x = 8x^2#

#4x * -2 = -8x#

#-15 * 2x = -30x#

#-15 * -2 = 30#

When we combine this all together we get:
#8x^2 - 8x - 30x + 30#

We can still combine like terms by doing #-8x - 30x#:
#8x^2 - 38x + 30#

#-------------------#

Now let's look at #(3x-1)^2# and expand:
#(3x-1)(3x-1)#

#3x * 3x = 9x^2#

#3x * -1 = -3x#

#-1 * 3x = -3x#

#-1 * -1 = 1#

When we combine this all together we get:
#9x^2 - 3x - 3x + 1#

Then we combine like terms by doing #-3x-3x#:
#9x^2 - 6x + 1#

#------------------#

So the equation is now:
#y = 8x^2 - 38x + 30 - (9x^2 - 6x + 1)#

Let's distribute the negative sign:
#y = 8x^2 - 38x + 30 - 9x^2 + 6x - 1#

Finally, let's combine like terms again:
#y = color(red)(8x^2) quadcolor(magenta)(-quad38x) + color(blue)30 quadcolor(red)(-quad9x^2) + color(magenta)(6x) quadcolor(blue)(-quad1)#

So the final answer in standard form is:
#y = -x^2 - 32x + 29#
as it matches #y = ax^2 + bx + c#.

Hope this helps!