# FIND ALL ZEROS OF THE POLYNOMIAL

To find zeroes of the polynomial, we have different methods as follows.

i) Factoring

ii) Using synthetic division

iii)  Using rational root theorem

Problem 1 :

Find all of the rational zeros of

f(x) = 5x3 + 12x2 - 29x + 12

Solution :

f(x) = 5x3 + 12x2 - 29x + 12

an = 5 and a0 = 12

Factor of a0, values of p = ±1, ±2, ±3, ±4, ±6, ±12

Factor of avalues of q = ±1, ±5

p/q =±1, ±2, ±3, ±4, ±6, ±12, ±2/5, ±3/5, ±4/5±6/5, ±12/5

Using synthetic division,

1 is the solution. To find the other two solutions, let us solve this quadratic equation got as quotient.

5x2 + 17x - 12 = 0

5x2 + 20x - 3x - 12 = 0

5x(x + 4) - 3(x + 4) = 0

(5x - 3) (x + 4) = 0

x = 3/5 and x = -4

So, the solutions are -4, 1 and 3/5.

Problem 2 :

f(x) = 8x4 + 2x3 + 5x2 + 2x - 3

Solution :

Using rational root theorem, we find the possible roots for the given polynomial.

an = 8 and a0 = -3

Factor of a0, values of p = ±1, ±3

Factor of avalues of q = ±1, ±2, ±4, ±8

p/q =±1, ±1/2, ±1/4±1/8, ±3/2, ±3/4, ±3/8

Using remainder theorem,

f(1) = 8(1)4 + 2(1)3 + 5(1)2 + 2(1) - 3

= 8 + 2 + 5 + 2 - 3

≠ 0

f(-1) = 8(-1)4 + 2(-1)3 + 5(-1)2 + 2(-1) - 3

= 8 - 2 + 5 - 2 - 3

≠ 0

f(1/2) = 8(1/2)4 + 2(1/2)3 + 5(1/2)2 + 2(1/2) - 3

= 8(1/16) + 2(1/8) + 5(1/4) + 1 - 3

= 1/2 + 1/4 + 5/4 - 2

= (2 + 1 + 5 - 8)/4

= 0

So 1/2 is one of the solutions

The quotient = 8x3 + 6x2 + 8x + 6

Solving this quotient, we will get the remaining zeroes.

2x2(4x + 3) + 2(4x + 3) = 0

(2x2 + 2)(4x + 3) = 0

2(x2 + 1)(4x + 3) = 0

x = -3/4 and x = ±i

So, the solutions are 1/2, -3/4 and ±i

Problem 3 :

x3 + 4x2 - 25x - 28

Solution :

From the above synthetic division, it is clear that -1 is one of the solution.

The quotient = x2 + 3x - 28 = 0

(x + 7)(x - 4) = 0

x = -7 and x = 4

So, the solution are -1, 4 and -7.

Problem 4 :

x4 + 2x3 - 11x2 + 8x - 60

Solution :

Using rational root theorem, we find the possible roots for the given polynomial.

an = 1 and a0 = -60

Factor of a0, values of p = ±1, ±2, ±3, ±4, ±5, ±10,  ±12, ±15, ±20, ±30, ±60

Factor of avalues of q = ±1

p/q = ±1, ±2, ±3, ±4, ±5, ±10,  ±12, ±15, ±20, ±30, ±60

So, 3 is one of the solutions.

The quotient = x3 + 5x2 + 4x + 20 = 0

x2 (x + 5) + 4 (x + 5) = 0

(x2 + 4) (x + 5) = 0

x2 + 4 = 0 and x + 5 = 0

x = ±2i and x = -5

Problem 5 :

x3 + 6x2 + 4x + 24

Solution :

Using rational root theorem, we find the possible roots for the given polynomial.

an = 1 and a0 = 24

Factor of a0, values of p = ±1, ±2, ±3, ±4, ±6, ±8,  ±12, ±24

Factor of avalues of q = ±1

p/q = ±1, ±2, ±3, ±4, ±6, ±8,  ±12, ±24

x3 + 6x2 + 4x + 24 = 0

x2(x + 6) + 4(x + 6) = 0

(x2+ 4) (x + 6) = 0

x2+ 4 = 0 and x = -6

x= - 4

x = ±i

So, the solution -6, i and -i.

Problem 6 :

4x4 + 5x3 + 30x2 + 45 x - 54

Solution :

Using rational root theorem, we find the possible roots for the given polynomial.

So, one of the solutions is -2.

4x3 - 3x2 + 36x - 27 = 0

x2(4x - 3) + 9(4x - 3) = 0

(x2+ 9)(4x - 3) = 0

x2+ 9 = 0 and 4x - 3 = 0

x= - 9 and x = 3/4

x = ±3i

So, the solutions are -2, 3/4, 3i and -3i

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