Chapter 1.3, Problem 129E

a.

To determine

To verify: The algebraic expression $ab=\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$ .

Explanation of Solution

Given information:

The algebraic expression $ab=\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$ .

Formula used:

The special factoring formula for perfect square which is mathematically expressed as,

  ${\left(A+B\right)}^2=A^2+2AB+B^2$

Proof:

Consider the function, $ab=\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$ . The right hand side of the equation can be solved separately to verify the equation.

The right hand side of the equation is,

  $\begin{array}{c}\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]=\frac{1}{2}\left(\left(a^2+b^2+2ab\right)-\left(a^2+b^2\right)\right)\\ =\frac{1}{2}\left(a^2+b^2+2ab-a^2-b^2\right)\\ =\frac{1}{2}\left(2ab\right)\\ =ab\end{array}$

Since, left hand side and right hand side are equal, therefore, the algebraic expression $ab=\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$ is verified.

b.

To determine

To verify: The algebraic expression ${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=4a^2{b}^2$ .

Explanation of Solution

Given information:

The algebraic expression ${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=4a^2{b}^2$ .

Formula used:

The special factoring formula for difference of squares, which is mathematically expressed as,

  $A^2-B^2=\left(A+B\right)\left(A-B\right)$

Proof:

Consider the equation, ${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=4a^2{b}^2$ . The left hand side of the equation can be solved separately to verify the equation.

The left hand side of equation is,

  $\begin{array}{c}{\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=\left(a^2+b^2+a^2-b^2\right)\left(a^2+b^2-a^2+b^2\right)\\ =\left(2a^2\right)\left(2b^2\right)\\ =4a^2{b}^2\end{array}$

Since, left hand side and right hand side are equal, therefore, the algebraic expression ${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=4a^2{b}^2$ is verified.

c.

To determine

To verify: The algebraic expression $\left(a^2+b^2\right)\left(c^2+d^2\right)={\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$ .

Explanation of Solution

Given information:

The algebraic expression, $\left(a^2+b^2\right)\left(c^2+d^2\right)={\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$ .

Formula used:

The special factoring formula for perfect square which is mathematically expressed as,

  ${\left(A+B\right)}^2=A^2+2AB+B^2$

  ${\left(A-B\right)}^2=A^2-2AB+B^2$

Proof:

Consider the equation, $\left(a^2+b^2\right)\left(c^2+d^2\right)={\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$ . The right hand side of the equation can be solved separately to verify the equation.

The right hand side of the equation is,

  $\begin{array}{c}{\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2={\left(ac\right)}^2+2\left(ac\right)\left(bd\right)+{\left(bd\right)}^2+{\left(ad\right)}^2-2\left(ad\right)\left(bc\right)+{\left(bc\right)}^2\\ =a^2{c}^2+2acbd+b^2{d}^2+a^2{d}^2-2adbc+b^2{c}^2\\ =a^2{c}^2+b^2{d}^2+a^2{d}^2+b^2{c}^2\end{array}$

Group the terms to take out common factors and simplify it further as,

  $\begin{array}{c}{\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2=a^2{c}^2+b^2{d}^2+a^2{d}^2+b^2{c}^2\\ =\left(a^2{c}^2+a^2{d}^2\right)+\left(b^2{d}^2+b^2{c}^2\right)\\ =a^2\left(c^2+d^2\right)+b^2\left(d^2+c^2\right)\\ =\left(a^2+b^2\right)\left(c^2+d^2\right)\end{array}$

Since, left hand side and right hand side are equal, therefore, the algebraic expression $\left(a^2+b^2\right)\left(c^2+d^2\right)={\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$ is verified.

d.

To determine

To calculate: The factor of the expression $4a^2{c}^2-{\left(a^2-b^2+c^2\right)}^2$ .

Answer to Problem 129E

The factor of the expression $4a^2{c}^2-{\left(a^2-b^2+c^2\right)}^2$ is $a^2\left(2c^2-a^2+2b^2\right)+b^2\left(2c^2-b^2\right)-c^4$ .

Explanation of Solution

Given information:

The expression $4a^2{c}^2-{\left(a^2-b^2+c^2\right)}^2$ .

Formula used:

To factor out the common factor from a polynomial, find out the greatest common factor and express the polynomial as a product of the simpler ones.

Calculation:

Consider the given expression $4a^2{c}^2-{\left(a^2-b^2+c^2\right)}^2$ .

Recall that to factor out the common factor from a polynomial, find out the greatest common factor and express the polynomial as a product of the simpler ones.

So, $4a^2{c}^2-{\left(a^2-b^2+c^2\right)}^2$ can be written in simplified form as,

  $\begin{array}{c}4a^2{c}^2-{\left(a^2-b^2+c^2\right)}^2=4a^2{c}^2-\left(a^2-b^2+c^2\right)\left(a^2-b^2+c^2\right)\\ =4a^2{c}^2-a^2\left(a^2-b^2+c^2\right)+b^2\left(a^2-b^2+c^2\right)-c^2\left(a^2-b^2+c^2\right)\\ =4a^2{c}^2-a^2\cdot a^2+a^2\cdot b^2-a^2\cdot c^2+b^2\cdot a^2-b^2\cdot b^2+b^2\cdot c^2-c^2\cdot a^2+b^2\cdot c^2-c^2\cdot c^2\\ =4a^2{c}^2-a^4+a^2{b}^2-a^2{c}^2+b^2{a}^2-b^4+b^2{c}^2-a^2{c}^2+b^2{c}^2-c^4\end{array}$

Simplify it further as,

  $\begin{array}{c}4a^2{c}^2-{\left(a^2-b^2+c^2\right)}^2=4a^2{c}^2-a^4+a^2{b}^2-a^2{c}^2+b^2{a}^2-b^4+b^2{c}^2-a^2{c}^2+b^2{c}^2-c^4\\ =4a^2{c}^2-a^4+2a^2{b}^2-2a^2{c}^2-b^4+2b^2{c}^2-c^4\\ =\left(2a^2{c}^2-a^4+2a^2{b}^2\right)+\left(2b^2{c}^2-b^4\right)-c^4\\ =a^2\left(2c^2-a^2+2b^2\right)+b^2\left(2c^2-b^2\right)-c^4\end{array}$

Thus, the common factor of the expression $4a^2{c}^2-{\left(a^2-b^2+c^2\right)}^2$ is $a^2\left(2c^2-a^2+2b^2\right)+b^2\left(2c^2-b^2\right)-c^4$ .