Question

4(a+b)^2−23(a+b)(c+d)+15(c+d)^2​

Answer 1

Explanation:

`4(a+b)^2-23(a+b)(c+d)+15(c+d)^2`

To simplify this equation, we'll apply the next propriety:

`(x+y)^2=x^2+2xy+y^2`

Thus:

`4(a^2+2ab+b^2)-23(a+b)(c+d)+15(c^2+2cd+d^2)`

Now, we apply the distributed propriety:

`(a+b)(c+d)=ac+ad+bc+bd\\\\k(x+y)=kx+ky`

Thus:

`4(a^2+2ab+b^2)-23(ac+ad+bc+bd)+15(c^2+2cd+d^2)\\\\4a^2+8ab+4b^2-23ac-23ad-23bc-23bd+15c^2+30cd+15d^2\\\\\boxed{4a^2+4b^2+15c^2+15d^2+8ab-23ac-23ad+8bc+23bd}`

`\text{-B$\mathfrak{randon}$VN}`