find the value of b given that the polynomial f(x)=x^3+bx^2-4x-5 when divided by x-3 and has a remainder of -8.

Question

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Answer 1

Hi Zhane, to find the value of b, divide the polynomial by x - 3.

 

x³ + bx² - 4x - 5 / x-3

 

Gives:

 

x² + (3+b)x + (5+3b) with a remainder of 10+9b.

 

The remainder of 10 + 9b must equal the remainder given, or -8.

 

10 + 9b = -8

 

Subtract 10 from both sides:

 

9b = -18

 

Divide both sides by 9:

 

b = -2

 

Substitute for b to check:

 

x³ - 2x² - 4x - 5 / x-3

 

Gives:

 

x² + x - 1 with a remainder of -8.

 

Questions?

Answer 2

By the Remainder Theorem, the remainder when f(x) is divided by x-3 is f(3).  

 

So, we need to find b so that f(3) = -8.

 

    f(3) = 27 + 9b - 12 - 5 = -8

 

                          10 + 9b = -8

 

                                 9b = -18

 

                                   b = -2

 

 

Answer 3

If you divide any polynomial p(x) by (x - r), then the remainder is p(r).

 

p(x) = x³ + bx² - 4x - 5

 

We want to have p(3) = -8.

 

p(3) = 3³ + b·3² - 4·3 - 5 = 9b + 10

 

9b + 10 = -8

 

9b = -18

 

b = -2

 

Let's check this:

 

x³ - 2x² - 4x - 5

 

= x³ - 3x² + x² - 3x - x + 3 - 8

 

= x²(x - 3) + x(x - 3) - (x - 3) - 8

 

= (x² + x - 1)(x - 3) - 8

 

So it works.