**Rational root theorem :**

If the polynomial P(x) = a_{ n} x^{ n} + a_{ n – 1} x^{ n – 1} + ... + a_{ 2} x^{ 2} + a_{ 1} x + a_{ 0} has any rational roots, then they must be of the form

± (factor of a_{ 0}/factor of a_{ n)}

Let us see some example problems to understand the above concept.

**Example 1 :**

State the possible rational zeros for each function. Then find all rational zeroes

f (x) = 5x³ + 29x² + 19x − 5

**Solution :**

Here a_{ 0 }= 5 and a_{ n }= - 5

Factors of the coefficient of x³ (5) are ±1,±5

Factors of constant term (-5) are ±1,±5

If there is any rational root for the given cubic polynomial, it must be in the form of ( ±1/1, ±5/5 and ±1/5)

To check whether ±1/5 is the rational root of the cubic polynomial, we can use any of the methods like remainder theorem, synthetic division.

P(x) = 5x³ + 29x² + 19x − 5

Since we have to find all other roots we can use synthetic division.

1/5 is only one rational root of the given cubic polynomial. Generally the cubic polynomial will have three roots. Out of these three zeroes one root is rational and the other roots are real values.

We can get other two roots by factorizing the quadratic polynomial 5x² + 30x + 25.

Dividing the whole equation by 5, we get

x² + 6x + 5 = 0

(x + 1) (x + 5) = 0

x + 1 = 0 x + 5 = 0

x = -1 and x = -5

Hence the three roots are 1/5, -1 and -5

**Example 2 :**

State the possible rational zeros for each function. Then find all rational zeroes

f (x) = 4x³ - 9x² + 6x − 1

**Solution :**

Here a_{ 0 }= 4 and a_{ n }= - 1

Factors of the coefficient of x³ (4) are ±1,±4 and ±2

Factors of constant term (-1) are ±1

If there is any rational root for the given cubic polynomial, it must be in the form of ( ±1/1, ±1/4 and ±1/2)

Let us check whether 1/4 is the root of the given polynomial.

P(x) = 4x³ - 9x² + 6x − 1

P(1/4) = 4(1/4)³ - 9(1/4)² + 6(1/4) − 1

= 4(1/64) - 9(1/16) + (6/4) − 1

= (1/16) - (9/16) + (6/4) − 1

= (1 - 9 + 24 - 16)/16

= (25 - 25)/16

Hence 1/4 is one of the rational root of the given cubic polynomial. We can get other roots using synthetic division.

We can get other two roots by factorizing the quadratic polynomial 4x² - 8x + 4.

Dividing the whole equation by 4, we get

x² - 2x + 1 = 0

(x - 1) (x - 1) = 0

x - 1 = 0 x - 1 = 0

x = 1 and x = 1

Hence the three roots are 1, 1 and 1/4.

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