Question

Find the quotient of the following polynomial function using the reverse tabular method, then express the polynomial as a product of the two factors:

(x^3 + 7x^2 + 3x - 6) / (x - 1)

Answer 1

The quotient of the polynomial (x^3 + 7x^2 + 3x - 6) divided by (x - 1) is x^2 + 6x - 3, which can be done using the reverse tabular method. Therefore, the original polynomial can be expressed as (x - 1)(x^2 + 6x - 3).

Step-by-step explanation:

The student has asked to find the quotient of the following polynomial function using the reverse tabular method:

(x^3 + 7x^2 + 3x - 6) / (x - 1)

Then, express the polynomial as a product of two factors.

Firstly, we set up the reverse tabular method with the divisor (x-1) to the left and the dividend (x^3 + 7x^2 + 3x - 6) atop the columns. We fill in the coefficients of the dividend in descending order of powers of x. Since we're dividing by (x - 1), we'll bring down the leading coefficient of the dividend as is, which is the first term of the quotient:

• 
  
• 1

Next, we multiply this term by the divisor (x-1) and write the result underneath the next coefficient of the dividend. We then add the two numbers in that column to get the next coefficient of the quotient:

• 
  
• (1)*(x - 1) = x - 1
• 
  
• 7 + (-1) = 6

Repeat this process with the new coefficient:

• 
  
• (6)*(x - 1) = 6x - 6
• 
  
• 3 + (-6) = -3

And finally:

• 
  
• (-3)*(x - 1) = -3x + 3
• 
  
• -6 + 3 = -3

This gives us the complete quotient of the division:

x^2 + 6x - 3

Therefore, the original polynomial can be expressed as a product of two factors:

(x - 1)(x^2 + 6x - 3)

We've successfully used the reverse tabular method and found the quotient and the resulting factors of the given polynomial.