# Algebra 2 help!! 10 pts!!?

find all the zeros in the polynomial function
f(x) = x^4 + 4x^3 - 6x^2 -36x -27

f(x)= x^3 + 5x^2 - 4x - 20

write a polynomial of least degree that has real coeffecients, the given zeros and a leading coeffiecient of 1.
-2, -2, 3, -4i

f(x) = x3 + 5x2 -4x -20

f(2) = 23 +5(22) -4(2) -20

=8 + 20 - 8 - 20

=28 - 28

= 0

f(2) = 0 then x - 2  is a factor.

f(x) = (x - 2)(x2 + 7x +10)

f(x) = (x - 2)(x2 + 5x + 2x + 10)

f(x) = (x - 2)(x(x + 5) + 2(x + 5))

f(x) = (x - 2)(x + 2)(x + 5)

f(x) = 0 then (x - 2)(x + 2)(x + 5) = 0

x - 2 = 0

x - 2 + 2 =2

x + 0 = 2

x = 2

x + 2 = 0

Subtract 2 to each side

x + 2 - 2 = -2

x + 0 = -2

x = -2

x + 5 = 0

Subtract 5 from each side

x + 5 - 5 = -5

x + 0 = -5

x = -5.

f(x) = x4 + 4x3 - 6x2 - 36x - 27

f(x) = (x + 1)(x3 + 3x2 - 9x - 27)

f(x) = (x + 1)(x - 3)(x2 + 6x + 9)

f(x) = (x + 1)(x - 3)(x + 3)2          ((a + b)2 = a2 +2ab + b2 )

if f(x) = 0 then (x + 1)(x - 3)(x + 3)2 = 0

x+1 = 0 , x - 3 = 0 and x + 3 = 0 , x + 3 = 0

x + 1 = 0

Subtract 1 from each side

x + 1 - 1 = -1

x + 0 = -1

x = -1

x - 3 = 0

x - 3 + 3 = 3

x - 0 = 3

x  = 3

x + 3 = 0

Subtract  3 from each side

x + 3 - 3 = -3

x + 0 = -3

x = -3

x + 3 = 0

x = -3.

x = 1 , x = -2 , x = -2, x = 3 and x = -4i

x - 1 = 0 , x + 2 = 0 , x + 2 = 0 , x - 3 = 0 and x + 4i = 0

The polynomial function is the products of the factors

(x - 1)(x + 2)(x + 2)(x - 3)(x + 4i)

(x2 +2x - x - 2)(x + 2)(x - 3)(x + 4i)

(x2 + x - 2)(x2 -3x + 2x - 6)(x + 4i)

(x2 + x - 2)(x2 - x - 6)(x + 4i)

(x4 - x3 - 6x2 + x3 - x2 - 6x - 2x2 + 2x + 12)(x + 4I)

simplify

(x4 - 9x2 - 4x + 12)(x + 4i)

(x5 + 4ix4 - 9x3 -4x2(1 + 9i) + x(12 - 16i) + 48i.

There fore the least degree polynomial function :(x5 + 4ix4 - 9x3 -4x2(1 + 9i) + x(12 - 16i) + 48i.

1) The polynomial function f (x )  = x4 + 4x3 - 6x2 - 36x - 27

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation anxn + an  1xn – 1 + ... + a1x + a0 = 0, then p is a factor of a0 and q is a factor if an.

If p/q is a rational zero, then p  is a factor of 27 and is a factor of 1.

The possible values of p  are   ± 1, ± 3, ± 9, ± 27.

The possible values for q  are ± 1

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 3, ± 9, ± 27.

Make a table for the synthetic division and test possible real zeros.

 p/q 1 4 -6 -36 -27 1 1 5 -1 -37 -64 -1 1 3 -9 -27 0

Since f (- 1) = 0,   x  =   - 1 is a zero. The depressed polynomial is  x3 + 3x2 - 9x - 27 = 0.

If p/q is a rational zero, then p  is a factor of 27 and is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q  = ± 1, ± 3, ± 9, ± 27.

Make a table for the synthetic division and test possible real zeros.

 p/q 1 3 -9 -27 1 1 4 -5 -32 -1 1 2 -11 -16 3 1 6 9 0

Since f (3)  =  0,  x = 3 is a zero. The depressed polynomial is  x2 + 6x + 9.

edited Jul 1, 2014 by david

Contd......

Since the depressed polynomial of this zero,  x2 + 6x + 9 = 0, is quadratic, use the factorisation method to find the roots of the quadratic equation  x2 + 6x + 9 = 0.

x2 + 3x + 3x + 9 = 0

x ( x + 3) + 3 (x + 3) = 0

(x + 3 )(x + 3) = 0

(x + 3)2 = 0

Apply squre root on each side.

x = - 3

x = - 3 is a zero.

Zeros of the f (x)  = x4 + 4x3 - 6x– 36x - 27  are -3, -1 and 3.

2) Identify Rational Zeros

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

The polynomial function f (x ) = x3 + 5x2- 4x - 20

If p/q  is a rational zero, then p  is a factor of 20 and q  is a factor of 1.

The possible values of p  are   ± 1, ± 2,  ± 4, ± 5, ± 10 and  ± 20.

The possible values for q  are ± 1.

So, p/q  = ± 1, ± 2,  ± 4, ± 5, ± 10 and  ± 20.

Make a table for the synthetic division and test possible  zeros.

 p/q 1 5 -4 -20 1 1 6 2 -18 -1 1 4 -8 -12 2 1 7 10 0

Since f (2)  =  0,   x  =  2 is a zero. The depressed polynomial is   x2 + 7x + 10 = 0

Since the depressed polynomial of this zero, x2 + 7x + 10, is quadratic, use the factorisation method to find the roots of the quadratic equation

x2 + 7x + 10 = 0

x2 + 5x + 2x + 10 = 0

x ( x + 5) + 2( x + 5) = 0

(x + 5)(x + 2) = 0

Apply zero product property.

x + 5 = 0 and x + 2 = 0

x = - 5 and x = -2.

Zeros of f (x ) = x3 + 5x2- 4x - 20 are 2 , -2 and -5.