Question

y³+ 6y² + 11y + 6 factorise

Answer 1

$\underline { \blue { Factor \: Theorem : }}$

If p(x) is a polynomial of degree n ≥ 1 and 'a' is any real number, then

i) (x-a) is a factor of p(x) ,if p(a) = 0

ii) and it's converse "if (x-a) is a factor of a polynomial p(x) then p(a) = 0 .

$Let \:p(y) = y^{3} + 6y^{2} + 11y + 6$

$By \:trial ,\: we\:find \:that \: p(-1) = 0$

$p(-1) = (-1)^{3} + 6(-1)^{2} + 11(-1) + 6 \\= -1 + 6 -11 + 6 \\= 12 - 12 \\= 0$

$So, (y+1) \: is \: a \: factor \: of \: p(y)$

$Now , divide \: p(x) \:by \: (y+1) , we \:get$

y+1 | y³+6y²+11y+6| y²+5y+6

***** y³+ y²

_______________

********* 5y² + 11y

********* 5y² + 5y

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***************** 6y + 6

***************** 6y + 6

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Remainder (0)

_________________

$Now, p(y) = (y+1)(y^{2}+5y+6) \\= (y+1)(y^(2}+2y+3y+6 \\= (y+1)[ y(y+2)+3(y+2)] \\= (y+1)(y+2)(y+3)$

Therefore.,

$\red{Factors \: of \:y^{3} + 6y^{2} + 11y + 6} \\\green {= (y+1)(y+2)(y+3)}$

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Answer 2

Answer:

The above Answer is correct