f(x)=x^4+6x^3+7x^2-6x-8 rational zeros, and factor af(x) into linear factors

Question

Given the polynomial function f(x) find the rational zeros, then the other zeros ( that is solve the equation f(x)=0) and factor into linear factors.

Answer 1

f (x) = x⁴ + 6x³+ 7x² - 6x - 8

If p/q is a rational zero, then p is a factor of -8 and q is a factor of 1.

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

The possible values for q are ± 1.

So, p/q = ± 1,   ± 2, ± 4.

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

 

p/q	1	6	7	-6	-8
1	1	7	14	8	0

 

Since f(1) = 0, you know that x = 1 is a zero. The depressed polynomial is x³  +  7x² + 14x + 8

Factor x³  +  7x² + 14x + 8

x³  +  7x² + 14x + 8 = 0

The possible values of p are   ± 1, ± 2, ± 4

The possible values for q are ± 1.

So, p/q = ± 1, ± 2, ± 4

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

 

p/q	1	7	14	8
-1	1	6	8	0

 

Since f(-1) = 0, you know that x = -1is a zero. The depressed polynomial is  x² + 6x + 8

Factor x² + 6x + 8

x² + 6x + 8 = 0

x² + 2x + 4x + 8 = 0

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

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

Apply Zero Product Property

(x + 2) = 0 or (x + 4) = 0

x = -2 , x = -4

The zeros of this function are x = -4, -2, -1 and 1

Answer 2

f(x) = x^4+6x^3+7x^2-6x-8

By rational root therom

Since the leading cooefficient is 1 and constant is 8.

The possible rational zeroes are ±1,±2,±4,±8.

Now test the each of possibilities.

f(x) = x^4+6x^3+7x^2-6x-8

P(1) = (1)^4+6(1)^3+7(1)^2-6(1)-8 = 1+6+7-6-8 = 0

P(-1) = (-1)^4+6(-1)^3+7(-1)^2-6(-1)-8 = 1-6+7+6-8 = 0

P(2) = (2)^4+6(2)^3+7(2)^2-6(2)-8 is not equals to 0.

P(-2) = (-2)^4+6(-2)^3+7(-2)^2-6(-2)-8 = 16-48+28+12-8 = 0

P(4) = (4)^4+6(4)^3+7(4)^2-6(4)-8 is not equals to 0.

P(-4)= (-4)^4+6(-4)^3+7(-4)^2-6(-4)-8 = 256-384+112+24-8 = 0

P(8) = (8)^4+6(8)^3+7(8)^2-6(8)-8 is not equls to 0.

P(-8) = (-8)^4+6(-8)^3+7(-8)^2-6(-8)-8 is not equals to 0.

Rational zeroes of given polynomial is x = 1,-1,-2,-4.