Question

$question$
Given that the zeros of the cubic polynomial x³-6x²+3x+10 are the form of 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 given polynomial.​

Answer 1

$\pink{\boxed{\boxed{\boxed{❛Solution}}}}$

$\bf\underline{\underline{✯Given:}}$

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

Therefore,

a+ (a+b) + (a+2b) =

$\frac{ - constat \: term}{coefficent \: of \: {x}^{3} }$

=

$- \frac{10}{1}$

a(2) (a+2b) = -10

a( a+2b). = -5

.. (a+b= 5)

(2-b) (2+b) = -5

4- b² = -5

b² = 9

・when b = 3, a= 2-3 = -1

・when b = -3, a= 2-(-3) = 5

☆Thus, the values of a and b are (-1,3) or (5,-3).

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$\bf\underline{\underline{✯Also:}}$

a= -1 ;; a+b = -1+3 = 2

And,

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

as well as,

a= 5 ;; a+b = 5 + (-3) = 2

and,

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

Hence, the zeroes of the gives polynomial are -1, 2 and 5.

Answer 2

Answer:

$\huge \huge \bf {\: \pmb{Question}}$

Q. Given that the zeros of the cubic polynomial x³-6x²+3x+10 are the form of 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 given polynomial.

Given :-

• a, a+b and a+2b are roots of x³-6x²+3x+10

Required to find :-

• value of a and b and zeros of polynomial

$\huge \huge \bf {\: \pmb{Solution}}$

given that a, a+b and a+2b are roots of x³-6x²+3x+10

we know that sum of roots = $\frac{-cofficient~of~x²}{coffiecient ~of~x³ }$

= a + 2b + a + a + b = $\frac{-(-6)}{1}$

= 3a + 3b = 6

= 3 ( a + b ) = 6

= a + b = $\frac{6}{3}$

= a + b = 2 --- ( 1 )

= b = 2 - a

we know that product of roots = $\frac{-constant}{coffiecient ~of~x³ }$

( a + 2b )( a + b )a = $\frac{-10}{1}$

a + b + b ) ( a + b ) 2 = $\frac{-10}{1}$

substituting equation 1 here,

= ( 2 + b ) (2)a = -10

= ( 2 + b ) 2a = -10

= ( 4 - a )2a = -10

= 4a - a² = $\frac{-10}{5}$

= a² - 4a + 5 = 0

by splitting the middle term

= a² - 5a + a - 5 = 0

= a ( a - 5 ) + 1 ( a - 5 ) = 0

= ( a + 1 ) ( a - 5 ) = 0

→ a + 1 = 0

• a = -1

→ a - 5 = 0

• a = 5

→ when a = -1

= a + b = 2

= -1 + b = 2

= b = 2 + 1

= b = 3

• a = -1 , b = 3

→ when a = 5

= a + b = 2

= 5 + b = 2

= b = 2 - 5

= b = -3

• a = 5 , b = -3

Now, zeros of the given polynomial.

→ a = -1 & b = 3

• a = -1

• a+b = (-1) + 3 = 2

• a+2b = (-1) + 2 ( 3 ) = -1 + 6 = 5

→ a = 5 & b = -3

• a = 5

• a + b = 5 + (- 3 )= 2

• a + 2b = 5 + 2(-3) = 5 + (-6 )= -1