Use the rational zeros theorem to find all the real zeros of the polynomial function.

Question

Use the zeros to factor f over the real numbers.

f(x)=x^4+6x^3-11x^2-24x+28

What  are the real zeros? x=

Use the real zeros to factor f. f(x)=

Answer 1

The function f (x ) = x⁴ + 6x³ - 11x² - 24x + 28

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 28 and q is a factor of 1.

The possible values of p are   ± 1, ± 2, ± 4, ± 7, ± 14 and ± 28.

The possible values for q are ± 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2, ± 4, ± 7, ± 14 and ± 28.

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

p/q	1	6	- 11	- 24	28
-1	1	5	-16	-8	36
1	1	7	-4	-28	0

Since f (1) = 0,  x  = 1 is a zero. The depressed polynomial is  x³ + 7x² - 4x - 28 = 0.

 

Answer 2

Contd...

If p/q is a rational zero, then p is a factor of 28 and q is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2, ± 4, ± 7, ± 14 and ± 28.

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

p/q	1	7	-4	-28
1	1	8	4	-24
-1	1	6	-10	-18
2	1	9	14	0

Since f(2) =  0,  x = 2 is a zero. The depressed polynomial is  x² + 9x + 14 = 0

Solve the equation by factoring.

 x² + 7x + 2x + 14 = 0

x (x + 7) + 2(x + 7) = 0

(x + 7) (x + 2) = 0

x + 7 = 0 and x + 2 = 0

x = -7 and x = - 2

Real zeros of the function f (x ) = x⁴ + 6x³ - 11x² - 24x + 28 are at x = 1, 2, -2 ,-7.

Factoring of f (x) = (x - 1)(x - 2)(x + 2) (x + 7).