Question

find the quadratic polynomia with zeroes 2 and 1by3

Answer 1

Answer:

Given A quadratic polynomial with zeros −2 and 1/3.

x=-2 and x=1/3

x+2=0 and x-1/3=0

(x+2)(x-1/3)=0

x^{2} -\frac{1}{3} x+2x-\frac{2}{3} =0x

2

−

3

1

x+2x−

3

2

=0

x^{2} +\frac{5}{3}x-\frac{2}{3}=0x

2

+

3

5

x−

3

2

=0

3x^{2}+5x-2=03x

2

+5x−2=0

Hence polynomial is [/tex]3x^{2}+5x-23x

2

+5x−2

Hope it helps

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Answer 2

Step-by-step explanation:

Step-by-step explanation:

GIVEN :- α = -2 and β = 1/3

formula :- ax²+bx + c =0

solution- ax² + (α+β)x + (αβ)

ax² + ( -2+1/3)x + ( -2*1/3)

ax² + (-6 +1) x + (-2/3)

ax² +(-5)x +(-2/3)

== ax² - 5x -2/3