Question

verify that - 2 ,1 and 3 are the zeros of the cubic polynomial x cube minus 2 x square - 5 x + 6 and check the relationship between zeros of the zeros and the coefficients​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$given \: cubic \: polynomial \: is \\ \: {x}^{3} - 2 {x}^{2} - 5x + 6$

Verification of given zero -2

by putting -2 in place of x

${( - 2)}^{3} - 2 {( - 2)}^{2} - 5( - 2) + 6 \\ \\ = - 8 - 8 + 10 + 6 \\ \\ = - 16 + 16 \\ \\ = 0 \: \: \: \: \: \: verified$

verification of given zero 1

by putting 1 in place of x

${(1)}^{3} - 2 {(1)}^{2} - 5(1) + 6 \\ \\ = 1 - 2 - 5 + 6 \\ \\ = 7 - 7 \\ \\ = 0 \: \: \: verified$

verification of given zero 3

by putting 3 in place of x

${(3)}^{3} - 2 {(3)}^{2} - 5(3) + 6 \\ \\ = 27 - 18 - 15 + 6 \\ \\ = 33 - 33 \\ \\ = 0 \: \: \: \: verified$

NOW,, since zeroes are verified so ,

three zeroes of given cubic polynomial are

$\alpha = - 2 \\ \\ \beta = 1 \\ \\ \gamma = 3$

on comparison of given cubic polynomial with standard form of cubic polynomial ,I.e,

$a {x}^{3} + b {x}^{2} + cx + d$

we will get Coeficients:

a = 1 ; b = -2 ; c = -5 ; d = 6

AS WE KNOW,

$\alpha + \beta + \gamma = \frac{ - b}{a} \\ \\ (putting \: values \: in \: lhs \: ) \\ -2 + 1 + 3 = 2 \\ (putting \: values \: in \: rhs) \\ \frac{ - ( - 2)}{1} = 2 \\ lhs \: = rhs \: \\ verified$

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} \\ \\ (putting \: values \: in \: lhs) \\ ( - 2)(1) + (1)(3) + (3)( - 2) \\ = - 2 + 3 - 6 = - 5 \\ (putting \: values \: in \: rhs) \\ \frac{ - 5}{1} = - 5 \\ lhs = rhs\\ verified$

$\alpha \beta \gamma = \frac{ - d}{a} \\ \\ (putting \: values \: in \: lhs) \\ ( - 2)(1)(3) = - 6 \\ (putting \: values \: in \: rhs) \\ \frac{ - (6)}{1} = - 6 \\ \\lhs = rhs \\ verified$