The polynomial which is in the form
ax^{3} + bx^{2} + cx + d
is known as cubic polynomial.
By solving cubic polynomial will have three values of x. Those are x-intercepts. To find those values, we should factories it.
Get the factors using grouping method, get linear factors. Then equating each factors to zero, we will get three values as solutions.
3) Remainder theorem.
Problem 1 :
Solving x^{3} + 9x^{2} - x - 9 = 0 we get the following roots
(a) ± 1,-9 (b) ±1, ±9 (c) ±1, 9 (d) None
Solution :
So, factors are (x - 1)(x^{2} + 10x + 9).
x^{2} + 10x + 9 = (x + 1) (x + 9)
Equating each factor to zero, we get
x + 1 = 0 x - 1 = 0 x + 9 = 0
x = -1, 1 and 9.
Problem 2 :
The solution of the cubic equation x^{3} - 6x^{2} + 11 x - 6 = 0 is given by the triplet
(a) (-1, 1, -2) (b) (1, 2, 3) (c) (-2, 2, 3) (d) (0, 4, -5)
Solution :
So, the factors are (x - 1) (x^{2} - 5x + 6)
x^{2} - 5x + 6 = (x - 2) (x - 3)
Equating each factor to zero, we get
x - 1 = 0 x - 2 = 0 x - 3 = 0
x = 1, 2 and 3
Problem 3 :
For the given polynomial
x^{3} - 2x^{2} - 5x + 6
If 3 is one of the zeroes, find the others.
writing the left over as quadratic polynomial, we get
x^{2} + x - 2
To get the other factors,
x^{2} + x - 2 = 0
(x + 2) (x - 1) = 0
x = -2 and x = 1
So, the three zeroes are -2, 1 and 3.
Problem 4 :
Is x = 4 a root of the equation x^{3} − 6x^{2} + 9x + 1 = 0?
Solution :
Let p(x) = x^{3} − 6x^{2} + 9x + 1
To check if x = 4 is a factor, we will use remainder theorem. If the remainder is 0, then 4 is a factor otherwise it is not.
p(4) = 4^{3} − 6(4)^{2} + 9(4) + 1
p(4) = 64 − 96 + 36 + 1
p(4) = 101 - 96
p(4) = 5
So, 4 is not a root of the polynomial.
Problem 5 :
Find a polynomial function P(x) of degree 3 with real coefficients that satisfies the given conditions.
Zeros of -3, -1, and 4; P(2) = 5
Solution :
Zeroes of the cubic polynomial are -3, -1 and 4.
x = -3, x = -1 and x = 4
Factors are (x + 3)(x + 1) (x - 4).
Multiplying the factors, we get
p(x) = k (x + 3)(x + 1) (x - 4)
p(x) = k (x^{2} + 4x + 3)(x - 4)
p(x) = k (x^{3} - 4x^{2} + 4x^{2} - 16x + 3x - 12)
p(x) = k (x^{3} - 13x - 12)
P(2) = 5
From the given information, the cubic polynomial should satisfy the condition P(2) = 5
p(2) = k (2^{3} - 13(2) - 12)
5 = k(8 - 26 - 12)
5 = k(-30)
-5/30 = k
k = -1/6
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM