Polynomials degree

Question

Find a polynomial of lowest degree with integer coefficients such that 1/2, 2/3, and 2 are zeros

Answer 1

A polynomial may be written as (x-a)(x-b)(x-c)...(x-n)=0.  Each of the constants (a, b, c, ... n) represent zeros.  The product of the expression on the left side is zero (to match the right side of the equation) because of the Zero Product Property (anything multiplied by zero results in a zero).  Polynomials written this way are in what is called Factored Form.

 

So, if the zeros occur at 1/2, 2/3, and 2, the polynomial is:

       (x-1/2)(x-2/3)(x-2)=0

       (2x-1)(3x-2)(x-2)=0      [multiply factors (and right side) by 2 and by 3 rid of fractions]

 

To write the polynomial in Standard From (with integer coefficients), simply multiply these terms [note: use distributive principle { a(b+c)=ab+ac } and F-O-I-L ]:

     ( (2x-1)(3x-2) )(x-2) = 0

     ( (2x-1)((3x-2) )x - 2( (2x-1)(3x-2) ) = 0            [distribute]

     (6x² - 4x -3x + 2)x - 2(6x² - 4x - 3x + 2) = 0       [F-O-I-L]

     6x³ - 7x² + 2x  - 12x² + 14x - 4 = 0                    [distribute]

     6x³ -19x² + 16x - 4 = 0                              [collect like terms]

        [note that the coefficients are all integers]

Answer 2

Hi Tamika

 

The answer is:  (2X-1)(3X-2)(X-2)

 

Why?  Hints:  The polynomial has 3 zeros at the same time...then, the degree of the polynomial is given by the number of zeros. The lowest polynomial degree is 3.

 

Because of 1/2 is a zero of this polynomial, we have that (2X-1) is true. Proof:  If (2X -1) is a zero, then is also true:

2X-1 =0. It happens only if X = 1/2.

 

The same reasoning can be applied for the other zeros: 2/3 and 2. 

 

Nicolas Martin

New Math tutor in WyzAnt