Chapter 2.2, Problem 7GP

Solution

To evaluate: A polynomial function for given x , using synthetic substitution.

The required value of $f\left(-1\right)$ is $-10$ .

Given data: The given polynomial function is $f\left(x\right)=-2x^4-x^3+4x-5$ and $x=-1$ .

Concept used:

If $\left(x-a\right)$ is a factor of the cubic polynomial function $f\left(x\right)$ , then it can be written as:

  $f\left(x\right)=A{\left(x-a\right)}^3+B{\left(x-a\right)}^2+C{\left(x-a\right)}^1+R$ , where R is the remainder when $f\left(x\right)$ is divided by factor $\left(x-a\right)$ . Then substituting $x=a$ ,

   $\begin{array}{c}f\left(a\right)=A{\left(a-a\right)}^3+B{\left(a-a\right)}^2+C{\left(a-a\right)}^1+R\\ =A{\left(0\right)}^3+B{\left(0\right)}^2+C{\left(0\right)}^1+R\\ =R\end{array}$

R is obtained by synthetic substitution.

Calculations:

The given polynomial in standard form is $f\left(x\right)=-2x^4-1x^3+0x^2+4x-5$ .

Step1: First write the coefficients of polynomial function $f\left(x\right)$ in order of descending exponents and given value $-1$ of x as

  $x-\text{value}\to -1\stackrel{\text{coefficients}}{\overbrace{-2-104-5}}$

Step2: Bring down the leading coefficient and multiply it by x value. Write the product under second coefficient and add,

  $\begin{array}{l}x-\text{value}\to -1\stackrel{\text{coefficients}}{\overbrace{-2-104-5}}\\ \underset{¯}{\downarrow 2}\\ -21\end{array}$

Step3: Multiply the sum (1) under second coefficient by x -value and write it under third coefficient and add as:

  $\begin{array}{l}x-\text{value}\to -1\stackrel{\text{coefficients}}{\overbrace{-2-104-5}}\\ \underset{¯}{\downarrow 2-1}\\ 21-1\end{array}$

Step3: Multiply the sum (-1) under third coefficient by x -value and write it under next coefficient and add as:

  $\begin{array}{l}x-\text{value}\to -1\stackrel{\text{coefficients}}{\overbrace{-2-104-5}}\\ \underset{¯}{\downarrow 2-11}\\ 21-15\end{array}$

Step4: Multiply the sum (5) under fourth coefficient by x -value and write it under next coefficient that is constant and add as:

  $\begin{array}{l}x-\text{value}\to -1\stackrel{\text{coefficients}}{\overbrace{-2-104-5}}\\ \underset{¯}{\downarrow 2-11-5}\\ 21-15-10\end{array}$

Thus, synthetic substitution gives $f\left(-1\right)=-10$ .

Check by direct substitution:

  $\begin{array}{c}f\left(x\right)=-2x^4-1x^3+0x^2+4x-5\\ f\left(-1\right)=-2{\left(-1\right)}^4-1{\left(-1\right)}^3+0{\left(-1\right)}^2+4\left(-1\right)-5\\ =-2+1+0-4-5\\ =-10\end{array}$