Multiply the polynomials using the special product formulas. Express your answer as a single polynomial in standard form.

$(x-2y)}^{2$

Ben Shaver
2021-12-16
Answered

Multiply the polynomials using the special product formulas. Express your answer as a single polynomial in standard form.

$(x-2y)}^{2$

You can still ask an expert for help

Dabanka4v

Answered 2021-12-17
Author has **36** answers

Step 1

Given:

$(x-2y)}^{2$

To multiply the polynomial express in standard form

Step 2

Special Identity$(a-b)}^{2}={a}^{2}-2ab+{b}^{2$

a=x; b=2y

$(x-2y)}^{2}={\left(x\right)}^{2}-2\left(x\right)\left(2y\right)+{\left(2y\right)}^{2$

$={x}^{2}-4xy+4{y}^{2}$

Answer:$x}^{2}-4xy+4{y}^{2$

Given:

To multiply the polynomial express in standard form

Step 2

Special Identity

a=x; b=2y

Answer:

Barbara Meeker

Answered 2021-12-18
Author has **38** answers

The given expression is the square of a binomial. So, multiply the binomials using the special products formula $(x-a)}^{2}={x}^{2}-2ax+{a}^{2$ .

$(x-2y)}^{2}={x}^{2}-2\cdot 2y\cdot x+{\left(2y\right)}^{2$

Rewrite using the laws of exponents and simplify.

$x}^{2}-2\cdot 2y\cdot x+{\left(2y\right)}^{2}={x}^{2}-4xy+{2}^{2}{y}^{2$

$={x}^{2}-4xy+4{y}^{2}$

Therefore, the result is$x}^{2}-4xy+4{y}^{2$ .

Rewrite using the laws of exponents and simplify.

Therefore, the result is

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How does $7\mathrm{log}(8x)=7\mathrm{ln}8x$

I was working on some math homework with a program called scientific notebook. I was check that

I was writing something correctly.

The original equation is $(\mathrm{log}({x}^{4})+\mathrm{log}({x}^{5}))/\mathrm{log}(8x)=7$

I then converted it to $\mathrm{log}({x}^{(4+5)})=7\mathrm{log}(8x)$

I was expecting to get $\mathrm{log}((8x{)}^{7})$ when I entered the $7\mathrm{log}(8x)$ into the program to evaluate. Can someone please explain this to me?

I was working on some math homework with a program called scientific notebook. I was check that

I was writing something correctly.

The original equation is $(\mathrm{log}({x}^{4})+\mathrm{log}({x}^{5}))/\mathrm{log}(8x)=7$

I then converted it to $\mathrm{log}({x}^{(4+5)})=7\mathrm{log}(8x)$

I was expecting to get $\mathrm{log}((8x{)}^{7})$ when I entered the $7\mathrm{log}(8x)$ into the program to evaluate. Can someone please explain this to me?

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