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The factorization of given cubic polynomial 6x³ - 11x² - 12x + 5 is:
6x³ - 11x² - 12x + 5 = (x + 1)(x - [tex]\frac{5}{2}[/tex])(x - [tex]\frac{1}{3}[/tex])
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 ± [tex]\frac{p}{q}[/tex] 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:
±{ [tex]\frac{1}{1} ,\frac{1}{2} ,\frac{1}{3} ,\frac{1}{6} ,\frac{5}{1} ,\frac{5}{2} ,\frac{5}{3} ,\frac{5}{6}[/tex] }
= ±{ [tex]1 ,\frac{1}{2} ,\frac{1}{3} ,\frac{1}{6} ,5 ,\frac{5}{2} ,\frac{5}{3} ,\frac{5}{6}[/tex] }
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 - [tex]\frac{5}{2}[/tex])(x - [tex]\frac{1}{3}[/tex])
So, the factors of given cubic polynomial are:
⇒ 6x³ - 11x² - 12x + 5 = (x + 1)(x - [tex]\frac{5}{2}[/tex])(x - [tex]\frac{1}{3}[/tex])
Therefore, the factorization of given cubic polynomial 6x³ - 11x² - 12x + 5 is: 6x³ - 11x² - 12x + 5 = (x + 1)(x - [tex]\frac{5}{2}[/tex])(x - [tex]\frac{1}{3}[/tex])
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