Question

factorise
y^3 - 3y^2- 9y- 5​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The answer is (y+1)(y+1)(y-5)

That is (y+1)whole^2(y-5)

Step-by-step explanation:

To factorise  y^3-3y^2-9y-5

Step 1 = First find the factors of the last term

              That is plus minus 1 and plus minus 5

let's try y = -1(factor theorem)

= y^3-3y^2-9y-5

= -1^3-3(-1)^2-9(-1)-5

= -1-3+9-5

= -4+4

= 0

so, -1 is one of the zero of the polynomial y^3-3y^2-9y-5

y+1= 0

y= -1

so(y+1) is one of the factor

Step 2 = Then divide the polynomialy^3-3y^2-9y-5 by (y+1)

            After dividing the quotient comes

           y^2-4y-5

Step 3 = Then factorise the quotient by splitting the middle term

            After factorising the factors of the quotient comes

            (y+1)(y-5)

so the factors of the polynomial y^3-3y^2-9y-5 are

(y+1)(y+1)(y-5)

That is (y+1) whole^2(y-5)