## 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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