Question

find the zeros of the polynomial and verify the relationship between zeros and coefficient

x^2+5x+6​

Answer 1

Answer:

${x}^{2} + 5x + 6 \\ {x}^{2} + 6x - x + 6 \\ x(x + 6) - 1(x + 6) \\ (x - 1)(x + 6) \\ x - 1 = 0 \: \: \: \: \: x + 6 = 0 \\ x = 1 \: \: \: \: \: x = - 6 \\ we \: get \: two \: zeroes \: 1 \: and \: - 6 \\ sum \: of \: zeroes = \frac{ - b}{a} \\ \alpha + \beta = \frac{ - b}{a} \\ - 6 + 1 = \frac{ - 5}{1} \\ - 5 = - 5 \\ \\ product \: of \: zeroes \: = \frac{c}{a} \\ \alpha \beta = \frac{c}{a} \\ ( 6)(1) = \frac{6}{1} \\ 6 = 6$

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Answer 2

Answer:

x2+3x+2x+6

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

(x+2)(x+3)

x=-2,x=-3

relationship

alpha +beta=-b/a

-3+(-2)=-5

-5=-5

now

alphaxbeta =c/a

-3x-2=6

6=6

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