Question

Evaluate the polynomial function at x=-2 using direct and synthetic substitution f(x)=3x⁵-x³+6x²-x+1

​

Answer 1

-61

by both methods

Explanation:

Let's first see what the terms direct substitution and synthetic substitution mean.

Direct Substitution

In direct substitution, to find the value of a polynomial f(x) at any point x = k. we simply substitute the value of k into the polynomial function and solve for f(k)

Synthetic Substitution

In synthetic substitution we make use of the Remainder Theorem..
The Remainder Theorem states that when we divide a polynomial
`f(x)` by
`x-c`, the remainder
`R` equals
`f(c)`

Solving using direct substitution

The given polynomial is

`f(x) = 3x^5 - x^3 + 6x^2 - x + 1`
and we are asked to evaluate this function at
`x = - 2`

Using Direct Substitution,

`f(-2) = =3\left(-2\right)^5-\left(-2\right)^3+6\left(-2\right)^2-\left(-2\right)+1\\\\=3\left(-32\right)-\left(-8\right)+6\left(-2\right)^2-\left(-2\right)+1\\\\=- -96 + 8 + 6(4) + 2 + 1\\=-96 + 8 + 24 + 2 + 1\\\\= -61\\\\`

Using direct substitution we get
`\bold{f(-2)= -61}`

Using Synthetic Substitution
Since the remainder when
`f(x)` is divided by
`x - c` is
`f(c)` , we will divide the polynomial
`f(x) = 3x^5 - x^3 + 6x^2 - x + 1` by
`x -(-2) = x + 2` and find out the remainder which will give f(-2)

This is the technique used in synthetic division
Step 1. Write only the coefficients of
`x` in the dividend inside an upside-down division symbol. Write -2 as the divisor on the left of this row

`\begin{matrix}\texttt\:\:\:3\:\:\:0\:\:-1\:\:\:6\:\:-1\:\:\:1\\ \;\;\;\;\;\;\;\texttt{\:\:\:\:|\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\;\;\;\;\;\;\;\;\;\;\;\;\;}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix}`

Step 2
Carry down the leading coefficient as is to below the division symbol

`\begin{matrix}\texttt\:\:\:3\:\:\:0\:\:-1\:\:\:6\:\:-1\:\:\:1\\ \;\;\;\;\;\;\;\texttt{\:\:\:\:|\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\;\;\;\;\;\;\;\;\;\;\;\;\;}}\\ \texttt{\:\:\:\:\;\;\;\;3\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix}`

Step 3
Multiply the above carry-down value by -2 and carry that result to the next column:

`3\left(-2\right)=-6`

`\begin{matrix}\texttt\:\:\:3\:\:\:\:\:\:\:0\:\:-1\:\:\:6\:\:-1\:\:\:1\\\texttt{\:\:|\ensuremath{\underline{\;\;\;\;\;\:\:\:-6\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}}\\\texttt{\:3\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix}`

Step 4

Add down the column

`\begin{matrix}\texttt\:\:\:\:\:\:-2\\\texttt{\:\:|\ensuremath{\underline{\;\;\;\;\;\:\:\:-6\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}}\\\texttt{\:3\:\:\:\:\:\:\:\:-6\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix}`

Step 5

Multiply the above carry-down value by -2 and carry that result to the next column:
(-6)(-2) = 12

2 | 3 0 -1 6 -1 1
| -6 12
-------------------------------------------------
3 -6

Step 6
Add down the column
-1 + 12 = 11

2 | 3 0 -1 6 -1 1
| -6 12
-------------------------------------------------
3 -6 11

Repeat this process for the other two coefficients

The final result is

2 | 3 0 -1 6 -1 1

| -6 12 -22 32 -62
-------------------------------------------------
3 -6 11 -16 31 -61

The last carry-down value s the remainder and the value of f(-2) = - 61