Question

Find a cubic polynomial whose zeros are 2,-3 &4

Answer 1

Given:

Zeroes of the polynomial are 2, -3, 4

To Find out :

A cubic polynomial ?

Solution:

As we know that :-

→ α + β + γ = -b/a

→ -3 + 2 + 4 = -b/a

→ 3/1 = -b/a ..... (i)

And

→ αβγ = -d/a

→ -3 × 2 × 4 = -d/a

→ -24/1 = -d/a .... (ii)

And

→ αβ + βγ + γα = c/a

→ -3×2 + 2×4 + 4×-3 = c/a

→ -6 + 8 - 12 = c/a

→ -10/1 = c/a .... (iii)

Now, From (i), (ii) and (iii), we get :-

• a = 1

• b = -3

• c = -10

• d = 24

As we know that :-

For a cubic polynomial :

→ ax³ + bx² + cx + d

→ (1)x³ + (-3)x² + (-10)x + (24)

→ x³ - 3x² - 10x + 24

Hence,

Hence, The cubic polynomial is x³ - 3x² - 10x + 24

Verification :-

→ x³ - 3x² - 10x + 24

→ x³ + 3x² - 6x² - 18x + 8x + 24

→ x²(x + 3) - 6x(x + 3) + 8(x + 3)

→ (x + 3)(x² - 6x + 8)

→ (x + 3)(x² - 2x - 4x + 8)

→ (x + 3)[x(x - 2) - 4(x - 2)]

→ (x + 3)(x - 2)(x - 4)

Zeroes are -

x + 3 = 0 and x - 2 = 0 and x - 4 = 0

x = -3, 2, 4

Here the zeroes comes as same as that is given in the question

Hence,

Verified.

Answer 2

$\large\underline\mathrm{Given:-}$

Zeroes of the polynomial ?

$\large\underline\mathrm{Solution}$

$\alpha + \beta + \gamma = - b \div a$

$\implies$ -3 + 2 + 4 = -b/a...(1)

and

$\alpha \beta \gamma = - d \div a$

$\implies$ -3 × 2 × 4 = -d/a...(2)

and

$\alpha \beta + \beta \gamma + \gamma \alpha = c \div a$

$\implies$ -3 × 2 + 2 × 4 - 4 × -3 = c/a

$\implies$ -6 + 8 - 12 = c/a

$\implies$ -10/1 = c/a

Now, from (1), (2) and (3), we get .

$\implies$ a = 1

$\implies$ b = -3

$\implies$ d = 24

For a cubic polynomial:

$\implies$ ax³ + bx² + cx + d

$\implies$ 1(x)³ + (-3)x² + (-10)x + 24

$\implies$ x³ - 3x² - 10x + 24

Hence, the cubic polynomial is x³ - 3x² - 10x + 24

$\implies$ x³ - 3x² - 10x + 24

$\implies$ x³ + 3x² - 6x² - 18x + 8x + 24

$\implies$ x²(x + 3) - 6x(x + 3) + 8(x + 3)

$\implies$ (x + 3)(x² - 2x - 4x + 8)

$\implies$ (x + 3) x(x - 2) -4(x + 2)

$\implies$ (x + 3)(x - 2)(x - 4)

Zeroes are = x = -3, 2, 4