Question

factorise the cubic polynomial Y3-6y2+11y-6

Answer 1

hope this help you
please plzzzzz mark as brainliest answer plzzzzz plzzzzzzzzzzzzz mark as brainliest answer plzzzzz please

Answer 2

The factors of $\bf {y}^{3} - 6 {y}^{2} + 11y - 6$ are $\bf \red{(y - 1)(y - 2)(y - 3)}$

Given:

• A cubic polynomial.
• ${y}^{3} - 6 {y}^{2} + 11y - 6$

To find:

• Factors of cubic polynomial.

Solution:

Concept to be used:

1. Find first factor by hit and trial method.
2. Divide the polynomial by factor and factorise quotient polynomial for other two factors.

Step 1:

Put y= 1

$= {(1)}^{3} - 6 {(1)}^{2} + 11(1) - 6$

or

$= 1 - 6 + 11 - 6$

or

$= 12 - 12$

or

$= 0$

Thus,

(y-1) is a factor of polynomial.

Step 2:

Divide the cubic polynomial by one of its known factor.

$y - 1 \: ) \: {y}^{3} - 6 {y}^{2} + 11y - 6 \: ( {y}^{2} - 5y + 6 \\ {y}^{3} - {y}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - \: \: \: \: \: \: \\ - 5 {y}^{2} + 11y \\ - 5 {y}^{2} + 5y \\ - - - - - - \\ 6y - 6 \\ 6y - 6 \\ - - - - - \\ 0 \\ - - - - - -$

Step 3:

Factorise the quotient polynomial.

${y}^{2} - 5y + 6$

or

$= {y}^{2} - 3y - 2y + 6$

or

$= y(y - 3) - 2(y - 3)$

or

$= (y - 3) (y - 2)$

Thus,

The factors of $\bf {y}^{3} - 6 {y}^{2} + 11y - 6$ are $\bf (y - 1)(y - 2)(y - 3)$

Learn more:

1) X3-7x-6. Factorise the polynomial

https://brainly.in/question/3903636

2) given that the zeros of the cubic polynomial x3-6x2+3x+10 are of the form a a+b a+2b for some positive real numbers, the...

https://brainly.in/question/40663192