Chapter 5, Problem 19PT

To determine

To calculate: The remaining factors of polynomial $x^3-4x^2+10x-12$ when it divided by $x-2$ .

Answer to Problem 19PT

The remaining factor of polynomial $x^3-4x^2+10x-12$ other than $x-2$ are $\left(x^2-2x+6\right)$ .

Explanation of Solution

Given information:

The polynomial is given as:

  $x^3-4x^2+10x-12$ and the dividend $x-2$ .

Formula used:

The synthetic division method,

Step 1: write the coefficients of the divided so that the degrees of the terms are in descending order. Then write the constant $r$ of the divisor $x-r$ in the box. Write the first coefficient down.

Step 2: Multiply the first coefficient by $r$ and write the product the second coefficient.

Step 3: Add the product and the second coefficient.

Step 4: Again, repeat Step 2 and 3 until reach a sum in last column. The number along the bottom row are the coefficient of the quotient. The final number is remainder.

The quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The Factor theorem: the binomial $x-r$ is a factor of the polynomial $p\left(x\right)$ if and only if

  $p\left(r\right)=0$

Calculation:

Consider the polynomial,

  $x^3-4x^2+10x-12$

The binomial $x-2$ is a factor of the polynomial if $2$ is a zero of the polynomials.

Using the factor theorem and synthetic division method as,

Step 1: Write the coefficient of dividend. The constant $r=2$ and bring the first coefficient 1 down.

  $\begin{array}{l}\overline{)2}1-410-12\\ \downarrow \\ \overline{1}\end{array}$

Step 2: Multiply the first coefficient by $r$ : $2\cdot 1=2$ .the product write under the second coefficient.

  $\begin{array}{l}\overline{)2}1-410-12\\ 2\\ \overline{1\nearrow }\end{array}$

Step 3: now write the sum of product and the second coefficient: $-4+\left(2\right)=-2$ .

  $\begin{array}{l}\overline{)2}1-410-12\\ 2\\ \overline{1-2}\end{array}$

Step 4: Multiply the sum,2 by $r:2\cdot -2=-4$ .The product value write under the next coefficient and add: $10-4=6$ .Now multiply the sum , $6$ by $r:6\cdot 2=12$ . The product value writes under the next coefficient and add: $12-12=0$ .

  $\begin{array}{l}\overline{)2}1-410-12\\ 2-412\\ \overline{1-2\nearrow 6\nearrow 0}\end{array}$

The remainder is 0 so, $x-2$ is a factor of the polynomial.

The polynomial can be factored as $\left(x-2\right)\left(x^2-2x+6\right)$ .The quotient $\left(x^2-2x+6\right)$ is in quadratic from but the factor are not in real number.

So, the factors of $x^3-4x^2+10x-12$ are $\left(x-2\right)\left(x^2-2x+6\right)$