Factorise x3 -6x2+11x-6?

Question

Answer

Let p(x) = x3-6x2+11x-6by trial, we find thatp(1)=(1)3 -6(1)2+11(1)-6=0by converse of  factor theorem:(x - 1) is a factor of p(x)Now, x3-6x2+11x-6=x2 (x-1) -5x (x-1) +6 (x-1) =(x-1) (x2 -5x +6)=(x-1) {x2 - 2x - 3x +6}=(x-1) {x(x-2)-3(x-2)}(x-1) (x-2) (x-3)

Answer

Let f(x) = x^3 - 6x^2 + 11x - 6. Let,X=1. Then,. F(X)=(1)^3-6×(1)^2×11×1-6. F(X)=1-6+11-6.F(X)=0.So,. X=1. (X-1) is a first factor. This type we can find second factor and third factor.For find scond factor take X=2 and for find third factor take X=3.X have power 3 that's why we can find three factors. So, The factors of this is (X-1)(X-2)(X-3).when you multiple this factors using identities you find X^3-6X^2+11X-6

Answer

Factorising x^3 - 6x^2 + 11x - 6

To factorise the given expression x^3 - 6x^2 + 11x - 6, we can use the factor theorem and synthetic division method.

Factor Theorem: If a polynomial P(x) has a factor (x - a), then P(a) = 0.

Synthetic Division: A method used to divide a polynomial by a linear expression.

Step 1: Finding the Possible Roots
To factorise the given expression, we need to find the possible roots. Since the coefficient of the highest power (1x^3) is 1, the possible rational roots can be found using the Rational Root Theorem.

The Rational Root Theorem states that if a polynomial has a rational root p/q, where p is a factor of the constant term and q is a factor of the leading coefficient, then p/q is a possible root.

The constant term is -6, and the leading coefficient is 1. Factors of -6 are ±1, ±2, ±3, and ±6. Factors of 1 are ±1.

Therefore, the possible rational roots are ±1, ±2, ±3, and ±6.

Step 2: Testing the Possible Roots
To determine which of the possible roots are actual roots, we can use synthetic division. We will test each possible root using synthetic division until we find a root that gives a remainder of 0.

Let's start with the possible root x = 1:

1 | 1 -6 11 -6
|__________
| 1 -5 6

The remainder is 6. Since it is not equal to 0, x = 1 is not a root.

Next, let's try x = -1:

-1 | 1 -6 11 -6
|__________
| -1 7 -4

Again, the remainder is not 0. So, x = -1 is not a root.

We continue testing the remaining possible roots until we find a root that gives a remainder of 0.

Step 3: Finding the Actual Root
After testing all the possible roots, we find that x = 2 is an actual root, as it gives a remainder of 0 when tested using synthetic division:

2 | 1 -6 11 -6
|__________
| 2 -8 6

The remainder is 0. Hence, x = 2 is an actual root.

Step 4: Factoring the Expression
Now that we have found an actual root, we can use synthetic division repeatedly to factorise the expression.

Using synthetic division with x = 2:

2 | 1 -6 11 -6
|__________
| 1 -4 14

The result is 1x^2 - 4x + 14.

Therefore, the expression x^3 - 6x^2 + 11x - 6 can be factorised as (x - 2)(x^2 - 4x +

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