EVALUATE POLYNOMIALS USING SYNTHETIC DIVISION

An elegant way of dividing a polynomial by a linear  polynomial was introduced by Paolo Ruffin in 1809. His method is known as synthetic division. It facilitates the division of a polynomial by a linear polynomial with the help of the coefficients involved.

Let us see how synthetic division can be used to explain the method of synthetic division with an example.

Let p(x) = x³ + 2x² - x - 4 be the dividend and q(x) = x + 2 be the divisor. We shall find the quotient s(x) and the remainder r, by proceeding as follows.

Step 1 :

Arrange the dividend and the divisor according to the descending powers of x and then write the coefficients of dividend in the first row (see figure). Insert 0 for missing terms.

Step 2 :

Find out the zero of the divisor.

x + 2 = 0

x = -2

Step 2 :

Put 0 for the first entry in the 2^nd row.

Complete the entries of the 2^nd row and 3^rd row as shown below.

Step 4 :

Write down the quotient and the remainder accordingly. All the entries except the last one in the third row constitute the coefficients of the quotient.

Thus, the quotient is x² - 1 and the remainder is –2.

In the above synthetic division. zero of the divisor is -2 and the remainder is -2.

This can be written as

p(-2) = -2

When we substitute the zero of the divisor -2 for x into p(x), the result is the remainder -2.

That is, evaluation of the polynomial p(x) for x = -2 is -2.

Example 1 :

If f(x) = x³ + x² - 7x - 3, then evaluate f(3).

Solution :

From the synthetic division above, we have

f(3) = 12

Example 2 :

If g(x) = 2x³ - 3x² - 3x + 2, then evaluate g(-1).

Solution :

From the synthetic division above, we have

g(-1) = 0

Example 3 :

If p(x) = x³ - 3x² - 10x + 24, then evaluate p(2).

Solution :

From the synthetic division above, we have

p(2) = 0

Example 4 :

If q(x) = 2x⁴ + x³ - 14x² - 19x + 6, then evaluate q(-1/2).

Solution :

From the synthetic division above, we have

q(-1/2) = 12

Write the function in the form f(x) = (x - k) q(x) + r for the given value of k, the demonstrate that f(k) = r.

Example 5 :

Given function is

f(x) = x³ - x² - 14x + 11

value of k is 4

Solution :

f(x) = (x - 4) q(x) + r

Dividing the polynomial by 4, we get

(x - 4)(x² + 3x - 2) + 3

Quotient = x² + 3x - 2 and

remainder = 3

Example 6 :

What is the value of k such that

(x³ − x² + kx − 30) ÷ (x − 5) has a remainder of zero?

a)  -14     b)  -2     c)  26     d)  32

Solution :

(x³ − x² + kx − 30) ÷ (x − 5)

The reaminder should be 0 when the given polynomial is divisible by the linear factor x - 5.

-30 + 5(k + 20) = 0

5(k + 20) = 30

k + 20 = 30/5

k + 20 = 6

k = 6 - 20

k = -14

So, the value of k is -14.

Example 7 :

The graph represents the polynomial function

f(x) = x³ + 3x² − x − 3

a. The expression f(x) ÷ (x − k) has a remainder of −15. What is the value of k?

b. Use the graph to compare the remainders of

(x³ + 3x² − x − 3) ÷ (x + 3) and (x³ + 3x² − x − 3) ÷ (x + 1)

Solution :

a)  By observing the graph, we know that one of the points is (-4, -15). So, dividing the polynomial by -4, we get the remainder as -15. Comparing with x - k and x - 4, the value of k is 4.

b) 

Observing the graph, -3 and -1 are the x-intercepts. The output is 0. So, the remainders of dividing the given polynomials by (x + 3) and (x + 1), we get the remainder as 0.

Example 8 :

The volume V of the rectangular prism is given by

V = 2x³ + 17x² + 46x + 40

Find an expression for the missing dimension

Solution :

V = 2x³ + 17x² + 46x + 40

Volume of rectangular prism = length x width x height

By observing the figure above, 

length = ?

width = x + 4 and 

height = x + 2

Dividing the cubic polynomial by the linear factor x + 2, we get 

2x² + 13x + 20 as quotient and 0 as remainder.

= 2x² + 13x + 20

= 2x² + 8x + 5x + 20

= 2x(x + 4) + 5(x + 4)

= (2x + 5)(x + 4)

So, the length of the rectangular prism is 2x + 5.

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