Question

Subtract 8x^2+3xy+4y^28x 2 +3xy+4y 2 8, x, squared, plus, 3, x, y, plus, 4, y, squared from 12x^2+3xy-5y^212x 2 +3xy−5y 2 12, x, squared, plus, 3, x, y, minus, 5, y, squared. Your answer should be a polynomial in standard form.

Answer 1

GIVEN :

The polynomials are $8x^2+3xy+4y^2$ and $12x^2+3xy-5y^2$.

TO FIND :

The subtraction of the given polynomial $8x^2+3xy+4y^2$ from the polynomial $12x^2+3xy-5y^2$.

SOLUTION :

Given polynomials are $8x^2+3xy+4y^2$ and $12x^2+3xy-5y^2$.

Now we can subtract $8x^2+3xy+4y^2$ from $12x^2+3xy-5y^2$.

Solving we get,

$12x^2+3xy-5y^2-(8x^2+3xy+4y^2)$

By using the Distributive Property :

a(x+y+z)=ax+ay+az

$=12x^2+3xy-5y^2-1(8x^2)+(-1)3xy+(-1)4y^2$

$=12x^2+3xy-5y^2-8x^2-3xy-4y^2$

$=4x^2+0xy-9y^2$

$=4x^2-9y^2$

∴ The answer to the given subtraction in standard form of the polynomial is $4x^2-9y^2$

By simplifying $4x^2-9y^2$

$=2^2x^2-3^2y^2$

$=(2x)^2-(3y)^2$

By using the Algebraic Identity:

$a^2-b^2=(a+b)(a-b)$

$=(2x+3y)(2x-3y)$

∴ $12x^2+3xy-5y^2-(8x^2+3xy+4y^2)=(2x+3y)(2x-3y)$