Question

Q2. Show that -1, 3 and 6 are zeroes of the polynomial p(x) = x 3 – 8x2 + 9x + 18. Also verify the relationship between the zeroes and the coefficients of p(x).

Answer 1

Given :

• Zeros of polynomial area -1, 3 and 6
• p(x) = x³ - 8x² + 9x + 18

According to the question,

➞ p(x) = x³ - 8x² + 9x + 18

➞ p(-1) = (-1)³ - 8(-1)² + 9(-1) + 18

➞ p(-1) = -1 - 8 - 9 + 18

➞ p(-1) = -9 - 9 + 18

➞ p(-1) = -18 + 18

➞ p(-1) = 0

➞ p(x) = x³ - 8x² + 9x + 18

➞ p(3) = (3)³ - 8(3)² + 9(3) + 18

➞ p(3) = 27 - 8(9) + 27 + 18

➞ p(3) = 27 - 72 + 45

➞ p(3) = -45 + 45

➞ p(3) = 0

➞ p(x) = x³ - 8x² + 9x + 18

➞ p(6) = (6)³ - 8(6)² + 9(6) + 18

➞ p(6) = 216 - 8(36) + 54 + 18

➞ p(6) = 216 - 288 + 72

➞ p(6) = -72 + 72

➞ p(6) = 0

★ Verifying the relationship between the zeros and the coefficients of p(x) :-

• α + β + γ = -b/a

➝ (-1) + 3 + 6 = -(-8)/1

➝ - 1 + 9 = 8

➝ 8 = 8

• αβ + βγ + γα = c/a

➝ (-1) × 3 + 3 × 6 + 6 × (-1) = 9/1

➝ -3 + 18 - 6 = 9

➝ 15 - 6 = 9

➝ 9 = 9

• αβγ = -d/a

➝ (-1) × 3 × 6 = -18/1

➝ -18 = -18

.°. Hence,verified.

Answer 2

Solution :-

Putting x = -1

$\sf (-1)^3 - 8(-1)^2+ 9(-1) + 18=0$

-1 - 8 + (-9) + 18 = 0

-9 + -9 + 18 = 0

-18 + 18 = 0

0 = 0

Putting x = 3

(3)³ - 8(3)² + 9(3) + 18 = 0

27 - 8(9) + 27 + 18 = 0

72 - 72 = 0

0 = 0

Putting x = 6

(6)³ - 8(6)² + 9(6) + 18 = 0

216 - 8(36) + 54 + 18 = 0

288 - 288 = 0

0 = 0

· V E R I F I C A T I O N :

-1 + 3 + 6 = -(-8)/1

-1 + 9 = 8/1

8 = 8

-1 × 3 + 3 × 6 + 6 × -1  = 9/1

-3 + 3 × 6 + -6 = 9

-9 + 18 = 9

9 = 9

-1 × 3 × 6 = -18/1

-1 × 18 = -18

-18 = -18