Question

y^3 -7y+6 please give solution step by step

Answer 1

$p(x )= y {}^{3} - 7y + 6$

Now put 1 in eq to find its factor.

$(1 ) {}^{3} - 7(1) + 6$

$7 - 7 = 0$

Since it satisfied the equTion so (y - 1) is the factor .

Now divide the whole equTion by (y - 1)

$(y {}^{3} - 7y + 6) \div (y - 1)$

On dividing above equations we get

$y {}^{2} - 6$

$(y - 1)(y {}^{2} - 6)$

hence to find zeroes equate the above factors with 0

y - 1 = 0

y = 1

$y {}^{2} - 6 = 0$

$y = + \sqrt{6} \: \: \: \: and \: \: - \sqrt{6}$

$hence \: zeroes \: are \: 1 \: \: + \sqrt{3 } \: \: \: and \: \: \: - \sqrt{3}$

Thanks