Question

Form a cubic polynomial whose zeroes are 2, 1/3 and 1/2.

Answer 1

Answer:

6x³ - 17x² + 11x - 2

Step-by-step explanation:

The cubic polynomial in general form is:

• ax³+bx²+cx+d = (x-x1)(x-x2)(x-x3)

zeroes are

• x1=2, x2=1/3, x3=1/2

then we can get the polynomial as:

• (x-2)(x-1/3)(x-1/2)=0
• (x-2)(3x-1)(2x-1)=0
• (3x²-x-6x+2)(2x-1)=0
• (3x²-7x+2)(2x-1)=0
• 6x³-3x²-14x²+7x+4x- 2=0
• 6x³ - 17x² + 11x - 2= 0

So the required polynomial is:

• 6x³ - 17x² + 11x - 2

Answer 2

Answer:

f(x) = 6x³ - 17x² + 11x - 2

Step-by-step explanation:

Given the zeros x = 2, x = $\frac{1}{3}$, x = $\frac{1}{2}$ then the factors ar

(x - 2), (x - $\frac{1}{3}$ ), (x - $\frac{1}{2}$ )

By equating the fractional factors to zero we can express them as integer factors, that is

x - $\frac{1}{3}$ = 0 ( multiply through by 3 )

3x - 1 = 0

x - $\frac{1}{2}$ = 0 ( multiply through by 2 )

2x - 1 = 0

Thus the factors are (x - 2), (3x - 1), (2x - 1)

The polynomial is then the product of the factors.

f(x) = (x - 2)(3x - 1)(2x - 1) ← expand last 2 factors using FOIL

     = (x - 2)(6x² - 5x + 1) ← distribute

     = 6x³ - 5x² + x - 12x² + 10x - 2 ← collect like terms

f(x) = 6x³ - 17x² + 11x - 2 ← cubic polynomial