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

2 Answers
Loweena Gonasegaran ยท Stefan V.
Aug 30, 2017

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

Alan N.
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

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