Question

find a quadratic polynomial whose sum and. product of the zeroes are _8\3and4\3 also find the zeroes of the polynomial by factorization ​

Answer 1

Solution :-

Sum of Zeroes = -8/3.

Product of Zeroes = 4/3.

We know that ;

• Polynomial = a[x² - (α + β)x + αβ]

-> a [x² - (-8/3)x + 4/3]

-> a [x² + 8/3x + 4/3]

Put the value of a = 3.

-> 3 [x² + 8/3 x + 4/3]

-> 3x² + 8x + 4

Its Zeroes are :-

-> 3x² + 8x + 4

Split the middle term by middle term splitting method.

-> 3x² + 6x + 2x + 4

-> 3x (x + 2) + 2 (x + 2)

-> (3x + 2) (x + 2)

• (3x + 2 = 0)

-> 3x = - 2

-> x = -2/3

• (x + 2 = 0)

-> x = - 2

Hence, the Polynomial is 3x² - 8x + 4 ;

and, Its zeroes are - 2 & - 2/3.

Answer 2

• We have to find a quadratic polynomial whose sum and product of the zeroes are $\dfrac{-8}{3}$ and $\dfrac{4}{3}$ and it's zeros also by factorization.

=> Sum of zeros = $\alpha \: + \: \beta$ = $\dfrac{-8}{3}$

=> Product of zeros = $\alpha \beta$ = $\dfrac{4}{3}$

______________________________

Now.. we have to form a quadratic equation.

We know that

• x² - (Sum of zeros) x + (Product of zeros)

Put the known values in above formula.

=> x² - $(\dfrac{-8}{3})$ x + $\dfrac{4}{3}$

Solve by taking LCM i.e. 3.

=> $\dfrac{3 {x}^{2} \: + \: 8x \: + \: 4}{3}$ = 0

=> 3x² + 8x + 4

______________________________

To find zeros of the above equation (i.e. 3x² + 8x + 4). We have to solve it by factorization method.

=> 3x² + 8x + 4 = 0

=> 3x² + 6x + 2x + 4 = 0

=> 3x(x + 2) +2(x + 2) = 0

=> (3x + 2) (x + 2) = 0

• 3x + 2 = 0

=> 3x = - 2

=> x = $\dfrac{-2}{3}$

• x + 2 = 0

=> x = - 2

______________________________

$\dfrac{-2}{3}$ and - 2 are the zeros.

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