polynomial with 3 being a zero

Question

For the polynomial below, 3 is a zero.

 

 h(x)=x³+3x²-16x-6

 

Express h(x) as a product of linear factors.

Answer 1

To express h(x) as a product of linear factors, we need to factor h(x) completely.

 

 

According to the rational root theorem, the possible roots of h(x) are

c = ±1, ±2, ±3, ±6

 

where  (x - c) is a possible factor.

 

We can use synthetic division to find the factor. The quotient when we perform the division must have a remainder of zero.

 

Since 3 is root, the remainder must be zero when we divide h(x) by 3.  Using synthetic division to find the other factor.

 

  

3 | 1     3    -16      -6

            

            3     18       6

__________________

     1     6       2       0

 

 

The quotient is     x² + 6x + 2

Reminder is the last digit, which is zero.

 

 

h(x) = (x - 3)(x² + 6x + 2)

 

 

Use the quadratic formula on the factor we just found.

 

x = (-6 ± √(36 - 4(2))) / 2

 

x = (-6 ± √(28)) / 2

 

x = (-6 ± 2√7) / 2

 

x = -3 ± √7

 

x = -3 - √7          and         x = -3 + √7

 

 

h(x) = (x - 3)(x + 3 + √7)(x + 3 - √7)

 

 

Since each factor has a degree of 1, they are linear.