Chapter 1.3, Problem 137E

  $\left(a\right)$

To determine

To calculate: The factors of the polynomial $x^4+x^2-2$ .

Answer to Problem 137E

The polynomial $x^4+x^2-2$ is factorized as $\left(x^2+2\right)\left(x-1\right)\left(x+1\right)$ .

Explanation of Solution

Given information:

The polynomial $x^4+x^2-2$ .

Formula used:

To factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

The difference of square of two numbers a and b is $a^2-b^2=\left(a+b\right)\left(a-b\right)$ .

Calculation:

Consider the polynomial $x^4+x^2-2$ .

Recall that to factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

Apply it,

  $\begin{array}{c}{x}^4+x^2-2=x^4+2x^2-x^2-2\\ =x^2\left(x^2+2\right)-1\left(x^2+2\right)\\ =\left(x^2+2\right)\left(x^2-1\right)\end{array}$

Recall that the difference of square of two numbers a and b is $a^2-b^2=\left(a+b\right)\left(a-b\right)$ .

Apply it,

  $\begin{array}{c}{x}^4+x^2-2=x^4+2x^2-x^2-2\\ =x^2\left(x^2+2\right)-1\left(x^2+2\right)\\ =\left(x^2+2\right)\left(x^2-1\right)\\ =\left(x^2+2\right)\left(x-1\right)\left(x+1\right)\end{array}$

Thus, the polynomial $x^4+x^2-2$ is factorized as $\left(x^2+2\right)\left(x-1\right)\left(x+1\right)$ .

  $\left(b\right)$

To determine

To calculate: The factors of the polynomial $x^4+2x^2+9$ .

Answer to Problem 137E

The polynomial $x^4+2x^2+9$ is factorized as $\left(x^2-2x+3\right)\left(x^2+2x+3\right)$ .

Explanation of Solution

Given information:

The polynomial $x^4+2x^2+9$ .

Formula used:

To factor a polynomial add and subtract the terms to make it a perfect square.

The square of sum of two numbers a and b is ${\left(a+b\right)}^2=a^2+b^2+2ab$ .

The difference of square of two numbers a and b is $a^2-b^2=\left(a+b\right)\left(a-b\right)$ .

Calculation:

Consider the polynomial $x^4+2x^2+9$ .

Recall that to factor a polynomial add and subtract the terms to make it a perfect square.

Add and subtract $4x^2$ ,

  $\begin{array}{c}{x}^4+2x^2+9=x^4+2x^2+4x^2+9-4x^2\\ =\left(x^4+6x^2+9\right)-4x^2\end{array}$

Now, factor the perfect square, square of sum of two numbers a and b is ${\left(a+b\right)}^2=a^2+b^2+2ab$ .

Simplify it further as,

  $\begin{array}{c}{x}^4+2x^2+9=x^4+2x^2+4x^2+9-4x^2\\ =\left(x^4+6x^2+9\right)-4x^2\\ ={\left(x^2+3\right)}^2-4x^2\end{array}$

Recall that the difference of square of two numbers a and b is $a^2-b^2=\left(a+b\right)\left(a-b\right)$ .

Apply it,

  $\begin{array}{c}{x}^4+2x^2+9=\left(x^2+3-2x\right)\left(x^2+3+2x\right)\\ =\left(x^2-2x+3\right)\left(x^2+2x+3\right)\end{array}$

Thus, the polynomial $x^4+2x^2+9$ is factorized as $\left(x^2-2x+3\right)\left(x^2+2x+3\right)$ .

  $\left(c\right)$

To determine

To calculate: The factors of the polynomial $x^4+4x^2+16$ .

Answer to Problem 137E

The polynomial $x^4+4x^2+16$ is factorized as $\left(x^2-2x+4\right)\left(x^2+2x+4\right)$ .

Explanation of Solution

Given information:

The polynomial $x^4+4x^2+16$ .

Formula used:

To factor a polynomial add and subtract the terms to make it a perfect square.

The square of sum of two numbers a and b is ${\left(a+b\right)}^2=a^2+b^2+2ab$ .

The difference of square of two numbers a and b is $a^2-b^2=\left(a+b\right)\left(a-b\right)$ .

Calculation:

Consider the polynomial $x^4+4x^2+16$ .

Recall that to factor a polynomial add and subtract the terms to make it a perfect square.

Add and subtract $4x^2$ ,

  $\begin{array}{c}{x}^4+4x^2+16=x^4+4x^2+4x^2+16-4x^2\\ =\left(x^4+8x^2+4^2\right)-4x^2\end{array}$

Now, factor the perfect square, square of sum of two numbers a and b is ${\left(a+b\right)}^2=a^2+b^2+2ab$ .

Simplify it further as,

  $\begin{array}{c}{x}^4+4x^2+16=x^4+4x^2+4x^2+16-4x^2\\ =\left(x^4+8x^2+4^2\right)-4x^2\\ ={\left(x^2+4\right)}^2-4x^2\end{array}$

Recall that the difference of square of two numbers a and b is $a^2-b^2=\left(a+b\right)\left(a-b\right)$ .

Apply it,

  $\begin{array}{c}{x}^4+4x^2+16=\left(x^2+4-2x\right)\left(x^2+4+2x\right)\\ =\left(x^2-2x+4\right)\left(x^2+2x+4\right)\end{array}$

Thus, the polynomial $x^4+4x^2+16$ is factorized as $\left(x^2-2x+4\right)\left(x^2+2x+4\right)$ .

  $\left(d\right)$

To determine

To calculate: The factors of the polynomial $x^4+2x^2+1$ .

Answer to Problem 137E

The polynomial $x^4+2x^2+1$ is factorized as ${\left(x^2+1\right)}^2$ .

Explanation of Solution

Given information:

The polynomial $x^4+2x^2+1$ .

Formula used:

To factor a polynomial add and subtract the terms to make it a perfect square.

The square of sum of two numbers a and b is ${\left(a+b\right)}^2=a^2+b^2+2ab$ .

The difference of square of two numbers a and b is $a^2-b^2=\left(a+b\right)\left(a-b\right)$ .

Calculation:

Consider the polynomial $x^4+2x^2+1$ .

Rewrite the polynomial as, ${\left(x^2\right)}^2+2\left(1\right)x^2+1^2$

Now, factor the perfect square, square of sum of two numbers a and b is ${\left(a+b\right)}^2=a^2+b^2+2ab$ .

  ${\left(x^2\right)}^2+2\left(1\right)x^2+1^2={\left(x^2+1\right)}^2$

Thus, the polynomial $x^4+2x^2+1$ is factorized as ${\left(x^2+1\right)}^2$ .