Chapter 2.4, Problem 61E

Solution

To prove: The upper and lower bound must be lie in the interval $\left[-5,4\right]$

Given information:

  $f\left(x\right)=x^4+2x^3-11x^2-13x+38$

Concept Used:

Divide a polynomial function $g\left(x\right)$ by $\left(x-k\right)$ , where $k<0$ , using synthetic division and all numbers are alternative sign at last line, then $k$ is an lower bound to the real roots of the equation $g\left(x\right)=0$ .

Divide a polynomial function $g\left(x\right)$ by $\left(x-k\right)$ , where $k>0$ , using synthetic division and all positive numbers at last line, then $k$ is an upper bound to the real roots of the equation $g\left(x\right)=0$ .

Calculation:

Setup the synthetic division for $x=-5$ ,

  $\begin{array}{cccccc}-5& 1& 2& -11& -13& 38\\ & \downarrow & -5& 15& -20& 165\\ & 1& -3& 4& -33& 203\end{array}$

Since the signs are alternative in bottom row, the $x=-5$ is lower bound of the real root.

Setup the synthetic division for $x=4$ ,

  $\begin{array}{cccccc}4& 1& 2& -11& -13& 38\\ & \downarrow & 4& 24& 52& 156\\ & 1& 6& 13& 39& 194\end{array}$

Since all the numbers in bottom row are positive, the $x=4$ is upper bound of the real root.

Hence proved.

To find: The rational zeros of the function $f\left(x\right)=x^4+2x^3-11x^2-13x+38$ .

  $\frac{p}{q}=\pm 1,\pm 2,\pm 19,\pm 38$

Given information:

  $f\left(x\right)=x^4+2x^3-11x^2-13x+38$

Concept Used:

Rational Zeros theorem states: If $g\left(x\right)$ is a polynomial with integer then $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.

Calculation:

  $f\left(x\right)=x^4+2x^3-11x^2-13x+38$

The leading coefficient is 1.

The factor of 1, $q=\pm 1$

The constant term is 38.

The factor of 38, $p=\pm 1,\pm 2,\pm 19,\pm 38$

Possible rational zeros are:

  $\frac{p}{q}=\pm 1,\pm 2,\pm 19,\pm 38$

To find: The rational zeros of the function $f\left(x\right)=x^4+2x^3-11x^2-13x+38$ .

  $\left(x-2\right)\left(x^3+4x^2-3x-19\right)$

Given information:

  $f\left(x\right)=x^4+2x^3-11x^2-13x+38$

Concept Used:

Rational Zeros theorem states: If $g\left(x\right)$ is a polynomial with integer then $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.

Calculation:

Possible rational zeros are:

  $\frac{p}{q}=\pm 1,\pm 2,\pm 19,\pm 38$

Now, check the factor of each possible zeros.

  $\begin{array}{cccccc}2& 1& 2& -11& -13& 38\\ & \downarrow & 2& 8& -6& -38\\ & 1& 4& -3& -19& 0\end{array}$

The first factor is $\left(x-2\right)$ and the other factor $\left(x^3+4x^2-3x-19\right)$ .

Now factor $\left(x^3+4x^2-3x-19\right)$ , this is a prime equation and not factor further.

The factor of $f\left(x\right)=x^4+2x^3-11x^2-13x+38$ is $\left(x-2\right)\left(x^3+4x^2-3x-19\right)$ .

To estimate: The irrational zeros of $f\left(x\right)=x^4+2x^3-11x^2-13x+38$ .

  $-3.02-0.45i\text{ and }-3.02+0.45i$

Given information:

  $f\left(x\right)=x^4+2x^3-11x^2-13x+38$

Calculation:

Now factor $\left(x^3+4x^2-3x-19\right)$ , this is a prime equation and not factor further.

Using calculator to estimate the zeros.

  $\begin{array}{l}{x}^3+4x^2-3x-19=0\\ x=2.04,-3.02-0.45i,-3.02+0.45i\end{array}$

The irrational zeros is about,

  $-3.02-0.45i\text{ and }-3.02+0.45i$

To find: The complete factor of $f\left(x\right)=x^4+2x^3-11x^2-13x+38$ .

  $\left(x-2\right)\left(x-\left(-3.02-0.45i\right)\right)\left(x-\left(-3.02+0.45i\right)\right)\left(x-2.04\right)$

Given information:

  $f\left(x\right)=x^4+2x^3-11x^2-13x+38$

Calculation:

Now factor $\left(x^3+4x^2-3x-19\right)$ , this is a prime equation and not factor further.

Using calculator to estimate the zeros.

  $\begin{array}{l}{x}^3+4x^2-3x-19=0\\ x=2.04,-3.02-0.45i,-3.02+0.45i\end{array}$

The irrational zeros is about,

  $-3.02-0.45i\text{ and }-3.02+0.45i$ ,

So, the factor of $f\left(x\right)=x^4+2x^3-11x^2-13x+38$ ,

  $\left(x-2\right)\left(x-\left(-3.02-0.45i\right)\right)\left(x-\left(-3.02+0.45i\right)\right)\left(x-2.04\right)$