HOW TO FIND THE MISSING VALUE OF CUBIC POLYNOMIAL IF ZEROES ARE GIVEN

How to Find the Missing Value of Cubic Polynomial if Zeroes are Given :

Here we are going to see, how to find the missing values of cubic polynomial if its zeroes are given.

How to Find the Missing Value of Cubic Polynomial if Zeroes are Given - Examples

Question 1 :

Find a number b such that 3 is a zero of the polynomial p defined by p(x) = 1 − 4x + bx² + 2x³

Solution :

p(x) = 1 − 4x + bx² + 2x³

If x = 3, then p(3)  =  0

p(3) = 1 − 4(3) + b(3)² + 2(3)³

0  =  1 - 12 + 9b + 2(27)

0  =  -11 + 9b + 54

0  =  43 + 9b

9b  =  -43

b  =  -43/9

Question 2 :

Find a number c such that −2 is a zero of the polynomial p defined by p(x) = 5 − 3x + 4x² + cx³

Solution :

p(x) = 5 − 3x + 4x² + cx³

If x = -2, then p(-2)  =  0

p(-2) = 5 − 3(-2) + 4(-2)² + c(-2)³

0  =  5 + 6 + 4(4) + c(-8)

0  =  27 - 8c

8c  =  27

c  =  27/8

Question 3 :

Find all choices of b, c, and d such that 1 and 4 are the only zeros of the polynomial p defined by p(x) = x³ + bx² + cx + d.

Solution :

p(x) = x³ + bx² + cx + d

The zeroes of the cubic polynomials are 1 and 4

x = 1, x = 4

By writing them as factors, we get (x - 1) and (x - 4). But we need to form a cubic equation. So, let us write (x - 1) twice or (x - 4) twice.

p(x)  =  (x - 1)² (x - 4)

  =  (x² - 2x + 1) (x - 4)

  =  x³ - 4x² - 2x² + 8x + x - 4

p(x)  =  x³ - 6x² + 9x - 4

p(x) = x³ + bx² + cx + d

b  =  -6, c  =  9 and d  =  -4

(or)

p(x)  =  (x - 4)² (x - 1)

  =  (x² - 8x + 16) (x - 1)

  =  x³ - 8x² + 16x - x² + 8x - 16

p(x)  =  =  x³ - 8x² - x²  + 16x + 8x - 16

p(x)  =  =  x³ - 9x² + 24x - 16

p(x) = x³ + bx² + cx + d

b  =  -9, c  =  24 and d  =  -16

Question 4 :

Find all choices of b, c, and d such that −3 and 2 are the only zeros of the polynomial p defined by p(x) = x³ + bx² + cx + d.

Solution :

p(x) = x³ + bx² + cx + d

The zeroes of the cubic polynomials are -3 and 2

x = -3, x = 2

By writing them as factors, we get (x + 3) and (x - 2). But we need to form a cubic equation. So, let us write (x + 3) twice or (x - 2) twice.

p(x)  =  (x + 3)² (x - 2)

  =  (x² + 6x + 9) (x - 2)

  =  x³ - 2x² + 6x² - 12x + 9x - 18

  =  x³ + 4x² - 3x - 18

p(x) = x³ + bx² + cx + d

b = 4, c = -3 and d = -18

p(x)  =  (x - 2)² (x + 3)

  =  (x² - 4x + 4) (x + 3)

  =  x³ + 3x² - 4x² - 12x + 4x + 12

  =  x³ - x² - 8x + 12

p(x) = x³ + bx² + cx + d

b = -1, c = -8 and d = 12

After having gone through the stuff given above, we hope that the students would have understood "How to Find the Missing Value of Cubic Polynomial if Zeroes are Given". 

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