What is the standard form of # y= (2x+14)(x+12)-(7x-7)^2#?

2 Answers

#y = -47x^2+ 136x +119 #

Explanation:

#y = (2x+14)(x+12)-(7x-7)^2#

#y=2x^2+24x+14x+168-(49x^2-98x+49)#

#y=2x^2+24x+14x+168-49x^2+98x-49#

#y=-47x^2+136x+119#

Aug 30, 2017

#y=-47x^2+136x+119#

Explanation:

The equation of a quadratic in standard form is: #y=ax^2+bx+c#

So, this question is asking us to find #a, b , c#

#y=(2x+14)(x+12) - (7x-7)^2#

It is probably simplier to break #y# in its two parts first.

#y = y_1 - y_2#

Where: #y_1 = (2x+14)(x+12)# and #y_2= (7x-7)^2#

Now, expand #y_1#

#y_1 = 2x^2+24x+14x+168#

#= 2x^2+38x+168#

Now, expand #y_2#

#y_2 = (7x-7)^2 = 7^2(x-1)^2#

#=49(x^2-2x+1)#

#= 49x^2-98x+49#

We can now simply combine #y_1 - y_2# to form #y#

Thus, #y= 2x^2+38x+168 -(49x^2-98x+49)#

#= 2x^2+38x+168 -49x^2+98x-49#

Combine coefficients of like terms.

#y = (2-49)x^2 + (38+98)x +(168-49)#

#y= -47x^2 + 136x +119# (Is our quadratic in standard form)

#a=-47, b=+136, c=+119#