Question

find a quadratic polynomial whose zeros exceed the zeros of x^2+5x+6 by number 2

Answer 1

from given polynomial x^2+5x+6 (since it is quadratic polynomial)

sum of roots = - b/a = -5 /1 = -5

product of the roots = c/a = 6/1 = 6

given for next polynomial the zeros are exceed by 2

Then Sum of roots (x+2)+(y+2) = -5
=> x+y = -5-4
therefore x+y = -9 or -9/1 ------ -b/a in second polynomial. ___ 1 eqn

product of roots (x+2)(y+2) = 6
xy+2x+2y+4 = 6
xy+2(x+y) = 6-4
xy+2(-9) = 2
xy=2+18
xy=20 or 20/1 -------- c/a in second polynomial ___ 2 eqn

so, from 1,2 eqns
a=1, b=9, c=20

therefore, required polynomial is x^2+9x+20=0

Answer 2

Answer:

The Solution Is In The Pic Above!

Step-by-step explanation:

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