Question

Factor the cubic polynomial 6x3 – 11x2 – 12x 5. use the rational root theorem and synthetic division

Answer 1

The factorization of given cubic polynomial 6x³ - 11x² - 12x + 5 is:  

6x³ - 11x² - 12x + 5 = (x + 1)(x - $\frac{5}{2}$)(x - $\frac{1}{3}$)

For given question,

We have been given the cubic polynomial 6x³ - 11x² - 12x + 5

We need to factorize given cubic polynomial.

By the rational roots theorem, any rational zero of f(x) is expressible in the form ± $\frac{p}{q}$ for integers p, q with p a divisor of the constant term 5 and q a divisor of the coefficient 6 of the leading term.

Factors of p = 5: 1, 5

Factors of q = 6: 1, 2, 3, 6

That means that the only possible rational zeros are:

±{ $\frac{1}{1} ,\frac{1}{2} ,\frac{1}{3} ,\frac{1}{6} ,\frac{5}{1} ,\frac{5}{2} ,\frac{5}{3} ,\frac{5}{6}$ }

= ±{ $1 ,\frac{1}{2} ,\frac{1}{3} ,\frac{1}{6} ,5 ,\frac{5}{2} ,\frac{5}{3} ,\frac{5}{6}$ }

We need to find the exact zeros of given cubic polynomial.

For x = 1,

6(1)³ - 11(1)² - 12(1) + 5 = -12

This means, x = 1 is not a zero of given cubic polynomial.

For x = -1,

6(-1)³ - 11(-1)² - 12(-1) + 5 = 0

This means, x = -1 is a zero of given cubic polynomial and (x + 1) is a factor.

To factorize given cubic polynomial we use synthetic division.

The synthetic division (6x³ - 11x² - 12x + 5) ÷ (x + 1) is as shown in following image.

⇒ 6x³ - 11x² - 12x + 5 = (x + 1)(6x² - 17x + 5)

The factors of above quadratic polynomial 6x² - 17x + 5 are:

⇒ 6x² - 17x + 5 = (x - $\frac{5}{2}$)(x - $\frac{1}{3}$)

So, the factors of given cubic polynomial are:

⇒ 6x³ - 11x² - 12x + 5 = (x + 1)(x - $\frac{5}{2}$)(x - $\frac{1}{3}$)

Therefore, the factorization of given cubic polynomial 6x³ - 11x² - 12x + 5 is:  6x³ - 11x² - 12x + 5 = (x + 1)(x - $\frac{5}{2}$)(x - $\frac{1}{3}$)

Learn more about the polynomial here:

brainly.com/question/16594008

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