Proving Polynomial Identities

a

We have a positive sign inside of the parentheses, we'll use the positive squared binomial.

(a + b)^{2} = a^{2} + 2 a b + b^{2} .

In our expression

(5 a b + 100)^{2}

the variables corresponds to

a = 5 a b

and

b = 100 .

Note that the entire term

5 a b

is to be raised by

2 .

(5 a b + 100)^{2}

(a + b)^{2} = a^{2} + 2 a b + b^{2}

(5 a b)^{2} + 2 \cdot 5 a b \cdot 100 + 10 0^{2}

(a b)^{m} = a^{m} b^{m}

5^{2} a^{2} b^{2} + 2 \cdot 5 a b \cdot 100 + 10 0^{2}

Simplify power and product

25 a^{2} b^{2} + 1000 a b + 10000

b

Again, we use the positive squared binomial, since there is a positive sign inside the parentheses.

(- 2 + x)^{2}

(a + b)^{2} = a^{2} + 2 a b + b^{2}

(- 2)^{2} + 2 \cdot (- 2) \cdot x + x^{2}

Simplify power and product

4 - 4 x + x^{2}

Use the negative squared binomial

We should also be able to expand the expression using the negative squared binomial if we change the order of terms inside the parentheses.

(- 2 + x)^{2}

CommutativePropAddCommutative Property of Addition

(x - 2)^{2}

(a - b)^{2} = a^{2} - 2 a b + b^{2}

x^{2} - 2 \cdot x \cdot 2 + 2^{2}

Simplify power and product

x^{2} - 4 x + 4

We got the same answer, but the terms ended up in a different order.

Proving Polynomial Identities 1.8 - Solution