Completely factor and find all zeros for each polynomial :
(1) x3 + 4x2 + 5x + 2
(2) x4- x3 + 14x2 - 16x - 32
(3) 5x3+29x2+19x-5
(4) 4x3 - 9x2 + 6x - 1
(5) 3x4- 10x3 - 24x2 - 6x + 5
(6) 3x3 + 9x2 + 4x + 12
Problem 1 :
x3 + 4x2 + 5x + 2
Solution :
Let p(x) = x3 + 4x2 + 5x + 2
By dividing the cubic polynomial by 1, we get 12 ≠ 0.
Dividing by -1, we get 0 as remainder. So, -1 is one of the solutions.
From the bottom row, we create a quadratic polynomial. By factoring this, we will get two factors.
x2+3x+2 = (x+1) (x+2)
x3+4x2+5x+2 = 0
(x+1) (x+1) (x+2) = 0
Equating each factor to zero, we get
x + 1 = 0, x + 1 = 0 and x + 2 = 0
x = -1, -1, -2
So, the solution is {-1, -1, -2}
Problem 2 :
x4- x3 + 14x2 - 16x - 32
Solution :
Let p(x) = x4- x3 + 14x2 - 16x - 32
By dividing the fourth degree polynomial by 1, we get -34≠0.
By applying the value of x as -1 and 2, we get the remainder as 0.
So, the factors are (x+1) and (x–2)
So far we got two roots of the given polynomial.
By solving,
x2 + 16 = 0
x2 = -16
x = √-16
x = ±4i
So, it has a imaginary root.
x4- x3 + 14x2 - 16x - 32
x + 1 = 0 and x - 2 = 0
x = -1, 2
So, the solution is {-1, 2, ±4i}
Problem 3 :
5x3+29x2+19x-5
Solution :
Let p(x) = 5x3+29x2+19x–5
By dividing the cubic polynomial by 1, we get 38≠0. Dividing by -1, we get 0 as remainder. So, -1 is one of the solution.
By factoring
quadratic polynomial 5x2 + 24x – 5, we get
5x2 + 24x - 5 = (5x - 1) (x + 5)
(5x - 1) (x + 5) = 0
Equating each factor to zero, we get
x = -1, 1/5, -5
So, the solution is {-1, 1/5, -5}
Problem 4 :
4x3 - 9x2 + 6x - 1
Solution :
Let p(x) = 4x3 - 9x2 + 6x - 1
By dividing the cubic polynomial by 1, we get 0 as remainder. So, 1 is one of the zeroes of the polynomial.
By factoring quadratic polynomial 4x2 - 5x + 1, we get
4x2 - 5x + 1 = (4x – 1) (x – 1)
The factors are (x – 1) (4x – 1) (x – 1)
Equating each factor to zero, we get
(x – 1) (4x – 1) (x – 1) = 0
x = 1, 1/4, 1
So, the solution is {1, 1/4, 1}
Problem 5 :
3x4- 10x3 - 24x2 - 6x + 5
Solution :
Let p(x) = 3x4- 10x3 - 24x2 - 6x + 5
By dividing the fourth degree polynomial by 1, we get -32≠0. So dividing by 5, we get 0 as remainder,
Both x = 5 and x = -1 are the two zeros of given polynomial.
By factoring quadratic polynomial 3x2 + 2x - 1, we get
3x2 + 2x - 1 = (3x – 1) (x + 1)
Equating each
factor to zero, we get
(3x – 1) (x + 1) = 0
x = 1/3, -1
So, the solution is {5, -1, 1/3, -1}
Problem 6 :
3x3 + 9x2 + 4x + 12
Solution :
Let p(x) = 3x3 + 9x2 + 4x + 12
By dividing the cubic polynomial by 1, we get 18≠0. Dividing by -3, we get 0 as remainder. So, -3 is one of the zeroes of the given polynomial.
By solving,
3x2 + 4 = 0
3x2 = -4
x = √-4/3
x = ±2/√3i
So, the solution
is {-3, ±2/√3i}.
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