Question

$two \: zeroes \: of \: a \: cubic \: polynimial \: a {x}^{3} + 3 {x}^{2} - bx + 6 \\ are \: - 1 \: and \: - 2.find \: the \: third \: zero \: and \: the \: values \: of \: a \: and \: b$
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Answer 1

let other zero be x

sum of zeroes = -3/a

-1 -2 +x = -3/a
-3 + x = -3/a

ALSO product of zeroes = -6/a

-1× -2× x = -6/a

x = 3/a

also x = -3/a +3
3/a = -3/a + 3

2/a = 1

a = 2

Answer 2

$\huge{\mathcal{Hi\: there!}}$

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p(x) =ax³+ 3x² - bx - 6

-1 and -2 are zeroes.

Therefore, p(-1) = 0 and p(-2) = 0

⇒ a( -1 )³ + 3( -1 )² - b( -1 ) - 6 = 0 and a × ( -2 )³ + 3 × (-2)² - b × (-2) - 6 = 0

-a + 3 + b - 6 = 0 and -8a + 12 + 2b - 6 = 0

-a + b = 3 and -8a + 2b = -6

-a + b = 3 -----------(1)

and

-4a + b = -3 ------------(2)

(1) - (2)

-a + b -(-4a + b) = 3 -(-3)

-a + b + 4a -b = 3+3

3a = 6

a = 6 / 3

$\boxed{\bold{a = 2}}$

-2 + b = 3 (substituting a=2 in (1) )

b = 3 + 2

$\boxed{\bold{b = 5}}$

p(x) = 2x³ + 3x² -5x - 6

Divide p(x) by ( x+1 ) by long division.

[Division is in the attachment]

We get the quotient as

2 x² + x - 6

= 2x² + 4x - 3x - 6 (splitting the middle term)

= 2x ( x+2 ) - 3 ( x+2 )

= ( x + 2 )( 2x - 3 )

To find the zero, put (x+2)(2x-3) = 0

Therefore, either (x+2) or (2x-3) = 0

If x+2 = 0, then x=-2, which is a zero

If 2x-3 = 0, then x=3/2, is a zero.

So, third zero = 3/2.

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Thanks for the question !

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