Question

Factorise the following y^3+2y^2+3y+6​

Answer 1

Answer:

           (y + 2) (y² + 3)

Step-by-step explanation:

An integer root would need to be a divisor of 6, so is one of ±1, ±2, ±3, ±6.

Putting a positive value for y will result in a positive value, so no root is positive.  Therefore the only candidates for integer roots are -1, -2, -3 and -6.

Start trying these one at a time:

• Putting y = -1 gives (-1)³+2(-1)²+3(-1)+6 = -1+2-3+6 ≠ 0, so -1 is not a root.
• Putting y = -2 gives (-2)³+2(-2)²+3(-2)+6 = -8+8-6+6 = 0, so -2 is a root!

By the Factor Theorem, y+2 is a factor.  Doing the division gives:

• y³ + 2y² + 3y + 6 = (y + 2)(y² + 3).

As y²+3 > 0 for all real y, this doesn't factorize any further (over the reals).

Answer 2

Solution :

Let p(y) be y³ + 2y² + 3y + 6

We shall now look for all the factors of 6. These are ±1, ±2, ±3,±6

By Hit and Trial Method :

When y = -2

★ p(-2) = (-2)³ + 2(-2)² + 3(-2) + 6

→ p(-2) = -8 + 2(4) - 6 + 6

→ p(-2) = - 8 + 8 - 6 + 6

→ p(-2) = - 14 + 14

→ p(-2) = 0

So, (y + 2) is factor of p(y)

Now,

• (y³ + 2y² + 3y + 6) ÷ (y + 2)

Refer to the attachment!

★ (y³ + 2y² + 3y + 6) ÷ (y + 2) = y² + 3

Hence,

$\dag \sf \: {y}^{3} + 2 {y}^{2} + 3y + 6 = (y + 2)( {y}^{2} + 3)$