Question

Please help me with 4/a, b, c and 5/b

Answer 1

Answer:

4:

a.) 49x^2-36y^2

b.) 9a^2-25b^2

c.) 1/25x^2-49

5: b.) -5x^2-4xy+4y^2

Step-by-step explanation:

Answer 2

Answers:

4a.) $49x^2-36y^2$

4b.) $9a^2-25b^2$

4c.) $\frac{1}{25}x^2-49$

5b.) $4y^2-4xy-5x^2$

Solution Steps:

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4a.) $\bold{(7x-6y)(7x+6y)}$:

 - Multiplication can be transformed into difference of squares using the rule: $(a-b)(a+b)=a^2-b^2$.

- So change the equation using the rule:

• $(7x)^2-(6y)^2$

- Expand $(7x)^2$ and $(6y)^2$:

• $(7x)^2=7^2x^2$

• $(6y)^2=6^2y^2$

- Calculate $7^2$ and $6^2$:

• $7^2=49$

• $6^2=36$

So the end equation would be: $49x^2-36y^2$.

4b.) $\bold{(3a+5b)(3a-5b)}$:

- Multiplication can be transformed into difference of squares using the rule: $(a-b)(a+b)=a^2-b^2$.

- So change the equation using the rule:

• $(3a)^2-(5b)^2$

- Expand $(3a)^2$ and $(5b)^2$ :

• $(3a)^2=3^2a^2$

• $(5b)^2=5^2b^2$

- Calculate $3^2$ and $5^2$:

• $3^2=9$

• $5^2=25$

So the end equation would be: $9a^2-25b^2$.

4c.) $\bold{(\frac{1}{5}x-7)}$ × $\bold{(\frac{1}{5}+7)}$:

- Multiplication can be transformed into difference of squares using the rule: $(a-b)(a+b)=a^2-b^2$.

- So change the equation using the rule:

• $(\frac{1}{5}x)^2-7^2$

- Expand $(\frac{1}{5}x)^2$:

• $(\frac{1}{5})^2x^2$

- Calculate $(\frac{1}{5})^2$ and $7^2$ :

• $(\frac{1}{5})^2=\frac{1}{25}$

• $7^2=49$

So the end equation would be: $\frac{1}{25}x^2-49$.

5b.) $\bold{(4x-y)(4x+y)+(2x-y)^2-(5x+2y)(5x-2y)}$:

- Consider $(4x-y)(4x+y)$. Multiplication can be transformed into difference of squares using the rule: $(a-b)(a+b)=a^2-b^2$.

- So change the equation using the rule:

• $(4x)^2-y^2$

- Expand $(4x)^2$:

• $(4x)^2=4^2x^2$

- Calculate $4^2$:

• $4^2=16$

- Use binomial theorem $(a-b)^2=a^2-2ab+b^2$ to expand $(2x-y)^2$:

• $4x^2-4xy+y^2$

- Combine $-y^2$ and $y^2$:

• $-y^2+y^2=0$

- Consider $(5x+2y)(5x-2y)$. Multiplication can be transformed into difference of squares using the rule: $(a-b)(a+b)=a^2-b^2$.

So change the equation using the rule:

• $(5x)^2-(2y)^2$

- Expand $(5x)^2$ and $(2y)^2$:

• $(5x)^2=5^2x^2$

• $(2y)^2=2^2y^2$

- Calculate $5^2$ and $2^2$:

• $5^2=25$

• $2^2=4$

- Combine $20x^2$ and $-25x^2$:

• $20x^2+(-25x^2)=-5x^2$

So the end equation would be: $4y^2-4xy-5x^2$.

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