Question

Given that the zeros of the cubic polynomial x³-6x+3x+10 are of the form a ,a+b and a+2b for some real number a and b ,find the values of a and b, as well as the zeros of the polynomial

Answer 1

Answer:

a = - 1 , b = 3 or a = 5, b = - 3 and - 1, 2, and 5 are the zeroes

Step-by-step explanation:

Given :

a, ( a + b ) , and ( a + 2b ) are the zeroes of the cubic polynomial x³ - 6x² + 3x + 10

Comparing x³ - 6x² + 3x + 10 with a'x³ + b'x² + c+ d we get

• a' = 1
• b' = - 6
• c = 3
• d = 10

Sum of zeroes = - b'/a

⇒ a + ( a + b ) + ( a + 2b ) = - (- 6 ) /1

⇒ 3a + 3b = 6

⇒ 3( a + b ) = 6

⇒ a + b = 6/3

⇒ a + b = 2 → ( 1 )

Product of zeroes = - d / a'

⇒ a( a + b )( a + 2b ) = - 10 / 1

It can be written as

⇒ a( a + b )( 2a + 2b - a ) = - 10

⇒ a( a + b ){ 2( a + b ) - a } = - 10

From Eq( 1 )

⇒ a( 2 ){ 2( 2 ) - a } = - 10

⇒ 2a( 4 - a ) = - 10

⇒ 8a - 2a² = - 10

⇒ 2a² - 8a - 10 = 0

Dividing on both sides by 2

⇒ a² - 4a - 5 = 0

Splitting the middle term

⇒ a² - 5a + a - 5 = 0

⇒ a( a - 5 ) + 1( a - 5 ) = 0

⇒ ( a + 1 )( a - 5 ) = 0

⇒ a + 1 = 0   OR   a - 5 = 0

⇒ a = - 1   OR   a = 5

Substituting the values of a in ( 1 )

Case 1 : When a =  - 1

⇒ - 1 + b = 2

⇒ b = 2 + 1

⇒ b = 3

Case 2 : When a = 5

⇒ 5 + b = 2

⇒ b = 2 - 5

⇒ b = - 3

Now, let's find the zeroes

Case 1 : When a = - 1 and b = 3

• a = - 1
• a + b = - 1 + 3 = 2
• a + 2b = - 1 + 2( 3 ) = - 1 + 6 = 5

Case 2 : When a = 5 and b = - 3

• a = 5
• a + b = 5 - 3 = 2
• a + 2b = 5 + 2( - 3 ) = 5 - 6 = - 1

Therefore the values of a and b are - 1 and 3 or 5 and - 3 respectively and the zeroes of the polynomial are - 1, 2, and 5.

Answer 2

Aɴꜱᴡᴇʀ

☞ A = 5 or (-1)

☞ B = (-3) or 3

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Gɪᴠᴇɴ

☆ Zeros of the polynomial x³ - 6x +3x +10 are of the form

◕ a

◕ a+b

◕ a+2b

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Tᴏ ꜰɪɴᴅ

✭ Value of a and b?

✭ Zeros of the polynomial?

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Sᴛᴇᴘꜱ

p(x) = x³ - 6x + 3x +10

❍ Given that a , a+b and a+2b are the zeros of p(x)

Then,

➤ a+(a+b)+(a+2b) = $\sf - \dfrac{(-6)}{1}$

➤ 3a + 3b = 6

➤ a + b = 3 ..... (i)

Sum of product of zeros at a time = $\sf\dfrac{coefficient \:of\:x}{coefficient\: of\: {x}^{3}}$

➼ a(a+b) + (a+b)(a+b) + a(a+2b) = $\sf\dfrac{3}{1}$

➼ a(a+b) + (a+b)(a+b) + b+a(a+b) + b = 3

➼ 2a + 2(2+b) + a(2+b) = 3 [using equation (i)]

➼ 2a + (2+2-a) + a(2+2-a) = 3 [using equation (i)]

➼ 2a + 8 - 2a + 4a - a² = 3

➼ - a² + 8 = 3 - 4a

➼ a² - 4a - 5 = 0

❍ Using factorisation method,

➳ a² - 5a + a - 5 = 0

➳ a(a-5) + 1(a-5) =0

➳ (a-5)(a+1) = 0

Equating each with 0,

✪ a - 5 = 0

✪ a = 5

_________

✪ a + 1= 0

✪ a = (-1)

_________

So of the a = (-1) ,then b = 3

And when a = 5 , then b = (-3)

Now lets substitute them and find the Zeros of the polynomial,

❍ When a = 5 and b = (-3)

➜ a , (a+b) and (a+2b) = 5 , (5-3) , (5+2(-3))

➜ 5 , 2 and -1

❍ When a = (-1) and b = 3

➜ a , (a+b) ,(a+2b) = (-1) , (-1+3) ,(-1+2(3))

➜ (-1) ,2 and 5

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