Question

factorise using the therom x3+6x3+11x+6

(all the 3 in the sum r cubes ontop of the x)

Answer 1

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Answer 2

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Correct question:

Factorise using the factor theorem:

x^3 + 6x^2 + 11x + 6.

Given:

• We have been given a cubic polynomial x^3 + 6x^2 + 11x + 6.

To Find:

• We need to find the factors of the given polynomial.

Solution:

Given f(x) = x^3 + 6x^2 + 11x + 6.

Putting the value of x as (-1) we have

f(-1) = (-1)^3 + 6(-1)^2 + 11(-1) + 6

= -1 + 6 - 11 + 6

= -12 + 12

= 0

Therefore, (x + 1) is a factor of x^3 + 6x^2 + 11x + 6.

Now, as we got (x + 1) as a factor of

x^3 + 6x^2 + 11x + 6, we need to divide this polynomial by (x+1).

[Division shown in attachment]

After dividing x^3 + 6x^2 + 11x + 6 by

(x + 1) we got quotient as x^2 + 5x + 6.

We can find the other factors by the method of splitting the middle term in

x^2 + 5x + 6.

So, we have

${x}^{3} + 6 {x}^{2} + 11x + 6 = (x + 1)( {x}^{2} + 5x + 6$

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

$= (x + 1){x(x + 2) + 3(x + 2)}$

$= (x + 1)(x + 2)(x + 3)$

Hence the factors of x^3 + 6x^2 + 11x + 6 are (x + 1), (x + 2) and (x + 3).