Question

find a quadratic polynomial whose sum and product respectively of the zeroes are –2 apon 3 and 4 apon 3 also find the zeroes of the polynomial by factaristion?​

Answer 1

Answer:

alpha+bheta=-8/3

alpha×bheta=4/3

Step-by-step explanation:

k(x^2-(alpha+bheta)+alpha×bheta

k(x^2-(-8/3)x+4/3)

k(3x^2+8/3x+4/3)

k(3x^2+8x+4/3)

3k(3x^2+8x+4/3)

the equation is 3x^2+8x+4.

3x^2+8x+4

=3x^2+6x+2x+4

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

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

x=-2,-2/3

Answer 2

Given

• $\text{-2/3 and 4/3 are the zeroes of the quadratic polynomial.}$

To Find

• Quadratic Polynomial
• Zeros of the polynomial by factorization

Solution

$Sum \: \: of \: \: zeros = \dfrac{ - 2}{3} + \dfrac{4}{3}$

$Sum \: \: of \: \: zeros = \dfrac{ - 2 + 4}{3}$

$\bf{Sum \: \: of \: \: zeros = \dfrac{ 2}{3} }$

$Product \: \: of \: zeros = \dfrac{ - 2}{ 3} \times \dfrac{4}{3}$

$\bf{Product \: \: of \: zeros = \dfrac{ - 8}{ 9} }$

we know that

• if a and ß are the zeros of Quadratic polynomial f(x) then the polynomial f(x) is given by

$~~~~~f(x) \sf{= k\{ x²-(\alpha+\beta)x + \alpha \beta \} }$

$~~~~~f(x) \sf{= k\{ x²-(Sum ~~of ~~zeros)x + Product \: \: of \: \: zeros \} }$

$~~~~~f(x) \sf{= x²- \bigg( \dfrac{2}{3} \bigg)x + \: \bigg( \dfrac{ - 8}{9} \bigg) }$

$~~~~~f(x) \sf{= x²- \dfrac{2x}{3} - \: \dfrac{8}{9} }$

• Hence the required Quadratic Polynomial f(x) is x²-2x/3 -8/9.

___________________

• Now finding the Zeros from Required Quadratic Polynomial.

$~~~~~f(x) \implies \sf{= x²- \dfrac{2x}{3} - \: \dfrac{8}{9} }$

• Multiply both sides by 9.

${~~~~ \implies \sf9 \bigg( x²- \dfrac{2x}{3} - \: \dfrac{8}{9} \bigg ) =9 \times 0}$

${~~~~ \implies \sf9x²- 6x - 8 =0}$

• Splitting Middle Term

${~~~~ \implies \sf9x²- 12x + 6x - 8 =0}$

${~~~~ \implies \sf3x(3x-4) +2( 3x - 4) =0}$

${~~~~ \implies \sf(3x-4) ( 3x + 2) =0}$

${~~~~ \implies \sf x = \dfrac{ - 2}{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \dfrac{4}{3} }$

• Zeros of Quadratic Polynomial be -2/3 and 4/3.