Question

in a cubic polynomial sum of zeros is 9 and product of zeros is 15 if zeros are in AP then find that cubic polynomial​

Answer 1

Solution

• Sum of zeros = 9
• Product of zeros = 15

→ α + ß + γ = 9

→ αßγ = 15

Since zeros are in A.P. hence we can take

• α = a - d
• ß = a
• γ = a + d

→ a - d + a + a + d = 9

→ 3a = 9

→ a = 3

→ (a - d)a(a + d) = 15

→ a³ - ad² = 15

→ 3³ - 3d² = 15

→ 27 - 15 = 3d²

→ 4 = d² → d = ± 2

Hence, Zeroes are 1, 3 and 5.

→ Polynomial = k[x³ - (α + ß + γ)x² + (αß + ßγ + αγ)x - (αßγ)]

→ Polynomial = k[x³ - (1 + 3 + 5)x² + (1 × 3 + 3 × 5 + 5 × 1)x - (1)(3)(5)]

→ Polynomial = k[x³ - 9x² + 23x - 15]

Answer 2

Answer:

$\large\boxed{\sf{ k({x}^{3} - 9 {x}^{2} + 23x - 15)\:\:,k\neq0}}$

Step-by-step explanation:

In a cubic polynomial, it is given that

Sum of zeroes = 9

Product of zeroes = 15

Let the zeroes be $\alpha,\beta\:and\;\gamma$

Also, zeroes are in AP

°.° Zeroes are in AP

Also, there are 3 zeroes in a cubic Polynomial.

.°. Let the zeroes be (a-d), a , (a+d)

.°. We have the relationship,

$\sf{ = > a - d + a + a + d = 9} \\ \\ \sf{ = > 3a = 9} \\ \\ \sf{ = > a = \frac{9}{3} }\\ \\ \sf{ = > a = 3 \: \: \: \: \: .........(i)}$

Again, we have the conditon,

$\sf{a(a - d)(a + d) = 15 }\\ \\ \sf{ = > a( {a}^{2} - {d}^{2} ) = 15 }\\ \\ \sf{= > 3( {3}^{2} - {d}^{2} ) = 15 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: from \: (i) }\\ \\ \sf{ = > 9 - {d}^{2} = \frac{15}{3} = 5} \\ \\ \sf{= > {d}^{2} = 9 - 5 = 4}\\ \\ \sf{= > {d}^{2} = {2}^{2} } \\ \\ \sf{ = > d = \pm2 \: \: \: \: \: \: \: \: .........(ii)}$

.°. From eqn (i) and (ii), we have,

$\sf{\alpha=a - d = 3- 2 = 1}\\\\\sf{\beta=a = 3}\\\\\sf{\gamma=a + d = 3 + 2 =5}$

.°. Zeroes are 1, 3 and 5

Thus, we have the required cubic Polynomial as

$\sf{=k( {x}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \beta \gamma + \gamma \alpha )x - ( \alpha \beta \gamma )) }\\ \\ \sf{ = k( {x}^{3} - (1 + 3 + 5) {x}^{2} + (1 \times 3 + 3 \times 5 + 5 \times 1)x - (1 \times 3 \times 5))} \\ \\ \sf{= k( {x}^{3} - 9 {x}^{2} + (3 + 15 + 5)x - 15 )}\\ \\ \sf{= k( {x}^{3} - 9 {x}^{2} + 23x - 15)}$

where, k is any non zero Integer.