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4(a+b)^2−23(a+b)(c+d)+15(c+d)^2​

User FuzzyChef
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1 Answer

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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}

User Ericb
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