Find all real zeros of the polynomial P(x)=x4−x3−19x2−x−20

Answer 1

x^4--x^3 -19x^2 -x -20

by descartes rule of signs there is 1 + real solutions, 1 or 3 - real solutions

and zero or 2 imaginary solutions

at least 2 real solutions

x f(x)

0 -20

-4 256 +64 -19(16) +4-20 = 0 2nd negative root is x=-4

1 1-1 -19-1-20 = -40

2 16 -8 -38 -2-20 = -58

3 81 - 27 -171 -3-20 = -140

4 256 -64 -19(16) -4-20 = --112-24 = -136

5 625 -125 -475 -5-20 = 0 x=5 is a positive real root

x=-4 and 5 are two real roots

5 1 -1 -19 -1 -20

1 4 1 4 0 means quotient= x^3 +4x^2 +x +4

-4 1 4 1 4

1 0 1 0 means final quotient = x^2 +1

x^2 =-1

x = + or - i are 2 imaginary roots

Answer 2

To find the zeros, set the polynomial equal to zero:

x⁴ − x³ − 19x² − x − 20 = 0

Using Descartes Rule of signs, there is one sign change meaning there is one positive real root (zero) so we'll start by looking for that zero.

Use the Rational Roots theorem to find all the possible rational roots (zeros):

Plus or minus the factors of 20 divided by the factors of 1: ±20, ±10, ±5, ±4, ±2, ±1.

Pick one of these to start with (I'll choose +1) and perform synthetic division:

1 | 1 -1 -19 -1 -20

|____1___0__19__18

1 0 19 18 -2

We got a remainder so +1 is not a zero.

Try +2:

2 | 1 -1 -19 -1 -20

|____2___2_-34_ -70

1 1 -17 -35 -90

We also got a remainder so +2 is not a zero but we see (by the Remainder Theorem, that the function value did not switch signs. We'll keep looking by trying +4:

4 | 1 -1 -19 -1 -20

|____4__12_-21__-88

1 3 -7 -22 -102

Again we got a remainder so +4 is not a zero but we see (by the Remainder Theorem), that the function value did not switch signs. We'll keep looking by trying +5:

5 | 1 -1 -19 -1 -20

|____5__20___5__20

1 4 1 4 0

We got a remainder of zero therefore x = 5 is a zero. The remaining cubic is x³ + 4x² + x + 4. We can factor it by grouping:

x³ + 4x² + x + 4 = 0

(x³ + 4x²) + (x + 4) = 0

x²(x + 4) + 1(x + 4) = 0

(x + 4)(x² + 1) = 0

Set each binomial equal to zero:

x = 4 = 0 so x = -4

x² + 1 = 0 so x² = -1 or x = ±√(-1) or x = ±i (this is not a real zero)

So the real zeros are x = -4 and x = 5

Answer 3

P(-4) = 0 and P(5) = 0

-4 | 1 -1 -19 -1 -20

-4. 20. -4. 20

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5 | 1. -5 1. -5 0

5 0 5

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1. 0 1 0

P(x) = (x + 4) (x - 5) (x² + 1)

x = -4 , 5