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if alpha and beta are zeroes of the polynomial f(x)= x^2+3x-10,then find a quadritic polynomial whose zeroes are 1/2alpha+beta and 1/2beta+alpha

Dear Student,We are given,f(x)=x2+3x-10 x2+5x-2x-10 = (x+5)(x-2)Therefore, α = -5 and β = 2Now, we have zeros of other polynomial as:-12α+β= 12*(-5)+2=-1212β+α=12*2-5=-4We know that formula of a polynomail with it's given zero's is k[ x2-sx+p]where s = sum of zeros and p = product of zeros Assuming k=1s=-12-4=-92p=-12*(-4)=2So, Required polynomial i: x2+92x+2=2x2+9x+4Regards

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First find the value of alpha and beta from f(x)=x^2+3x-10
x^2+5x-2x-10 = (x+5)(x-2), so from here we got the value for alpha and beta ,i.e., alpha = -5 and beta = 2
Now in the question zeros are given in the form of alpha and beta so fill the values and you will get your two zeroes.
After this there is a formula for making a polynomial from sum and product for zeros. Formula is x^2-sx+p
where s = sum of zeros and p = product of zeros. As we know the values for our zeroes , put those in this formula and hence you will get your answer.
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