Question

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

Answer 1

`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`

Answer 2

`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`