Question

If the zeros of the polynomial
$f(x) = {2x}^{3} - {15x}^{2} + 37x - 30$
are in A. P. (Arithmetic Progression), find them. ​

Answer 1

Answer : 2, 5/2, 3

Given polynomial : 2x³ - 15x² + 37x - 30 = 0

Zeros of polynomial be [ a - b ], [a], [ a + b ]

Comparing the given polynomial with ax³ + bx² + cx + d = 0, we get :

a = 2, b = -15, c = 37, d = -30

As we know,

Sum of zeros = -b/a

» (a - b) + a + (a + b) = - (-15)/2

» a = 5/2

Also, Product of zeroes = -d/a

» (a - b)(a)(a + b) = -(-30)/2

» a ( a² - b² ) = 15

.°. Substituting value of a in above equation,

(5/2) × [ ( 5/2 )² - b² ] = 15

» 25/4 - b² = 6

» 25/4 - 6 = b²

» 1/4 = b²

» b = ± 1/2

Therefore,

Zeroes of polynomial :

a - b = 5/2 - 1/2 = 4/2 = 2

a = 5/2

a + b = 5/2 + 1/2 = 6/2 = 3

Therefore,

Therefore, Zeroes of the polynomial are : 2, 5/2, 3.

Answer 2

Answer:

$\left( 3 \right), \left( \dfrac{5}{2} \right), \left( 2 \right)$

Step-by-step explanation:

We are required to find the zeroes of the polynomial $f(x)={2x}^{3}-{15x}^{2} +37x-30$

It is a cubic polynomial so it will have 3 roots.

The roots are in AP.

Let the roots to be (a-d), a, (a+d)

Sum of roots of a cubic polynomial is = $\dfrac{-(Co-effecient \ of \ x^{2})}{Co-efficient \ of \ x^{3}}$

$(a+d)+a+(a-d)=\dfrac{-(-15)}{2}$

$3a=\dfrac{15}{2}$

$a=\dfrac{5}{2}$

Product of roots of a cubic polynomial is = $\dfrac{-(Constant \ Term)}{Co-efficient \ of \ x^{3}}$

$a(a-d)(a+d)=\dfrac{-(-30)}{2}$

$a(a^{2}-d^{2})=15$

Substitute the value of a :-

$\dfrac{5}{2} \left[ \left( \dfrac{5}{2} \right)^{2} - d^{2} \bigg] = 15$

$\left[ \left( \dfrac{5}{2}\right)^{2} - d^{2} \bigg] = 6$

$\dfrac{25}{4} - d^{2} = 6$

$d^{2} = \dfrac{1}{4}$

$d=\sqrt{\dfrac{1}{4}}$

$d=\pm\dfrac{1}{2}$

So, the roots now would be:-

Case 1

If d is positive-

$\left( \dfrac{5}{2} - \dfrac{1}{2} \right), \left( \dfrac{5}{2} \right), \left( \dfrac{5}{2} + \dfrac{1}{2} \right)$

$(2), \left( \dfrac{5}{2} \right), (3)$

Case 2

If d is negative-

$\left( \dfrac{5}{2} - \dfrac{-1}{2} \right), \left( \dfrac{5}{2} \right),\left( \dfrac{5}{2} + \dfrac{-1}{2} \right)$

$(3), \left( \dfrac{5}{2} \right), (2)$

The roots remain the same in both the cases.

So, the roots are $(3), \left( \dfrac{5}{2} \right), (2)$