Question

(a-1)x^2-2bxy-(a+1)y^2​

Answer 1

Answer:

$\Large\boxed{ax^2-x^2-2bxy-ay^2-y^2}$

Step-by-step explanation:

Given:

(a-1)x²-2bxy-(a+1)y² (Solve to expand expression and distributive property)

To solve this problem, first you have to used the distributive property by expanding expressions.

Solutions:

First, you have to expand by using with distributive property.

$\Large\boxed{\textnormal{DISTRIBUTIVE PROPERTY FORMULA}}}$

$\displaystyle A(B+C)=AB+AC$

A=X²

B=A

C=1

Rewrite the whole problem down.

$\displaystyle X^2a-X^2*1$

$\displaystyle AX^2-1*X^2$

Multiply the numbers from left to right.

$\displaystyle X^2*1=X^2$

$\displaystyle AX^2-X^2$

$AX^2-X^2-2bxy-(a+1)y^2$

Expand solving with distributive property.

$\displaystyle -y^2(a+1)$

A=(-y²)

B=A

C=1

Change negative sign to positive sign.

$+(-K)=-K$

$-Ay^2-1*y^2$

Multiply.

Solve.

$\displaystyle Y^2*1=Y^2$

$\displaystyle\boxed{-ay^2-y^2}$

$\Large\boxed{ax^2-x^2-2bxy-ay^2-y^2}$

So, the correct answer is ax²-x²-2bxy-ay²-y².

Answer 2

Answer:

$\mathrm{Expand}\:\left(a-1\right)x^2-2bxy-\left(a+1\right)y^2:\quad ax^2-x^2-2bxy-ay^2-y^2$

Step-by-step explanation:

$\left(a-1\right)x^2-2bxy-\left(a+1\right)y^2$

$=x^2\left(a-1\right)-2bxy-y^2\left(a+1\right)$

$\black{\mathrm{Expand}\:x^2\left(a-1\right):}$

$x^2\left(a-1\right)$

$\gray{\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac}$

$\gray{a=x^2,\:b=a,\:c=1}$

$=x^2a-x^2\cdot \:1$

$=ax^2-1\cdot \:x^2$

$\gray{\mathrm{Multiply:}\:1\cdot \:x^2=x^2}$

$=ax^2-x^2$

$=ax^2-x^2-2bxy-\left(a+1\right)y^2$

$\black{\mathrm{Expand}\:-y^2\left(a+1\right):}$

$-y^2\left(a+1\right)$

$\gray{\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac}$

$\gray{a=-y^2,\:b=a,\:c=1}$

$=-y^2a+\left(-y^2\right)\cdot \:1$

$\gray{\mathrm{Apply\:minus-plus\:rules}}$

$\gray{+\left(-a\right)=-a}$

$=-ay^2-1\cdot \:y^2$

$\gray{\mathrm{Multiply:}\:1\cdot \:y^2=y^2}$

$=-ay^2-y^2$

$=ax^2-x^2-2bxy-ay^2-y^2$