Question

Verify that 1, -1 and – 3 are the zeroes of cubic polynomial x' + 3x - x -3 and check the

relationship between zeroes and coefficient​

Answer 1

$\huge\mathbb\red{ANSWER}$

$let \: \alpha \: be \: 1 \: and \: \beta \: be \: - 1 \: and \: \gamma \: = - 3$

$as \: \: we \: know \: sum \: of \: zeros \: \alpha + \beta + \gamma = \frac{ - b}{ + a} \\ \\ by \: putting \: value \: \\ \\ ( - 1) + ( - 3 )+ 1 = \frac{ - 3}{ + 1} \\ \\ 1 - 1 - 3 = \frac{ - 3}{ + 1} \\ \\ - 3 = - 3 \\ \\ lhs \: = \: rhs \: hence \: proved \:$

$as \: we know \: sum \: and \: product \: zero \: \alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a} \\ \\ ( - 1)1 + (- 3)( - 1) + 1( -1 ) = \frac{ - 1}{ + 1} \\ \\ - 1 + 3 - 1 = \frac{ - 1}{ + 1} \\ \\ - 1 = - 1 \\ \\ hence \: proved \:$

$as \: we \: know \: \alpha \beta \gamma = \frac{ - d}{a} \\ \\( - 1)(1) (- 3) = \frac{ - 3}{1} \\ \\ - 3 = - 3 \\ \\ 3 = 3$

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