What is the standard form of # y= (x-6)(4x+1)-(2x-1)(2x-2)#?

1 Answer
Jun 12, 2017

See a solution process below:

Explanation:

First, expand the terms in parenthesis by multiplying each set of individual terms in the left parenthesis by each set of individual terms in the right parenthesis.

#y = (color(red)(x) - color(red)(6))(color(blue)(4x) + color(blue)(1)) - (color(green)(2x) - color(green)(1))(color(purple)(2x) - color(purple)(2))# becomes:

#y = (color(red)(x) xx color(blue)(4x)) + (color(red)(x) xx color(blue)(1)) - (color(red)(6) xx color(blue)(4x)) - (color(red)(6) xx color(blue)(1)) - ((color(green)(2x) xx color(purple)(2x)) - (color(green)(2x) xx color(purple)(2)) - (color(green)(1) xx color(purple)(2x)) + (color(green)(1) xx color(purple)(2)))#

#y = 4x^2 + x - 24x - 6 - (4x^2 - 4x - 2x + 2)#

#y = 4x^2 + x - 24x - 6 - 4x^2 + 4x + 2x - 2#

We can next group like terms:

#y = 4x^2 - 4x^2 + x - 24x + 4x + 2x - 6 - 2#

Now, combine like terms:

#y = 4x^2 - 4x^2 + 1x - 24x + 4x + 2x - 6 - 2#

#y = (4 - 4)x^2 + (1 - 24 + 4 + 2)x + (- 6 - 2)#

#y = 0x^2 + (-17)x + (-8)#

#y = -17x - 8#

This is the standard form for a polynomial. However, the standard form for a linear equation, which this is, is: #color(red)(A)x + color(blue)(B)y = color(green)(C)#

Where, if at all possible, #color(red)(A)#, #color(blue)(B)#, and #color(green)(C)#are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

If this is what is desired we can convert as follows:

#color(red)(17x) + y = color(red)(17x) + -17x - 8#

#17x + 1y = 0 - 8#

#color(red)(17)x + color(blue)(1)y = color(green)(-8)#