Question

A quadrilateral polynomial whose zeroes are - 3 and 4 is

Answer 1

Answer:

Answer:Α = -3

Answer:Α = -3β = 4

Answer:Α = -3β = 4Now,

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a}

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a}

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1Therefore,

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1Therefore,b= -1 , a=1

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1Therefore,b= -1 , a=1Now,

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1Therefore,b= -1 , a=1Now, \alpha * \beta = \frac{c}{a}

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1Therefore,b= -1 , a=1Now, \alpha * \beta = \frac{c}{a} -12= \frac{c}{a}

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1Therefore,b= -1 , a=1Now, \alpha * \beta = \frac{c}{a} -12= \frac{c}{a} Therefore, c= -12.

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1Therefore,b= -1 , a=1Now, \alpha * \beta = \frac{c}{a} -12= \frac{c}{a} Therefore, c= -12.We know standard form of a quadratic polynomial = a x^{2} +bx+c

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1Therefore,b= -1 , a=1Now, \alpha * \beta = \frac{c}{a} -12= \frac{c}{a} Therefore, c= -12.We know standard form of a quadratic polynomial = a x^{2} +bx+cHence required polynomial= x^{2} -x-12

Answer:Α = -3β = 4Now, \alpha + \beta = \frac{-b}{a} -3+4 = \frac{-b}{a} \frac{-b}{a} = 1Therefore,b= -1 , a=1Now, \alpha * \beta = \frac{c}{a} -12= \frac{c}{a} Therefore, c= -12.We know standard form of a quadratic polynomial = a x^{2} +bx+cHence required polynomial= x^{2} -x-12Hope it helped. Do check my calculations! ;P

Answer 2

Answer:

$x^{2} -x-12$

Step-by-step explanation:

given: zeroes are -3, 4

=>α=-3     β=4

=>-b/a=α+β=4-3=1/1

=>b=-1, a=1

c/a=αβ=4(-3)=-12

∴ the Q.E. is $x^{2} -x-12$