Question

I really need help with this question! Please Help Precalculus?

Answer 1

The function F ( x ) = X⁵ - 4x⁴ + 4x³ + 2x² - 5x +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.

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation a_nxⁿ + a_n – 1x^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

If p/q is a rational zero, then p is a factor of 2 and q is a factor of 1.

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

The possible values for q are ± 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2 .

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

p/q	1	- 4	4	2	- 5	2
1	1	- 3	1	3	- 2	0

Since f(1) = 0, x = 1 is a zero. The depressed polynomial is  x⁴ -3x³ + x² + 3x - 2 = 0.

Now consider x⁴ -3x³ + x² + 3x - 2 = 0 .

If p/q is a rational zero, then p is a factor of -2 and q is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2 .

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

p/q	1	- 3	1	3	- 2
1	1	- 2	- 1	2	0

Since f(1) = 0, x = 1 is a zero. The depressed polynomial is  x³ -2x² - x + 2 = 0.

Now consider x³ -2x² - x + 2 = 0 .

If p/q is a rational zero, then p is a factor of 2 and q is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2 .

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

p/q	1	-2	- 1	2
1	1	-1	- 2	0

Since f(1) = 0, x = 1 is a zero. The depressed polynomial is  x² - x - 2 = 0.

Now consider x² - x - 2 = 0 .

x² - x - 2 = 0

x² - 2x + x - 2 = 0

x(x-2) +1(x-2) = 0 

(x-2)(x+1) = 0

x - 2 = 0 and x +1 = 0

x = 2 and x = -1 

So the solutions for the polynomial are -1,1,2 .

So first option is correct .