Question

Solve the following equation by factors

x³ - 6x² + 11x - 6 = 0​

Answer 1

Answer

• (x-3) , (x-2) & (x-1) are the roots of the equation

Explanation

Given

• p(x) = x³-6x²+11x-6

To Find

• The roots of the Equation

Solution

So here we shall use the trial and error method to find the first root [Here we see that 6 is the constant term here so we shall use their factors and see which one gives 0]

➝ P(x) = x³-6x²+11x-6

➝ P(3) = 3³ - 6(3)² + 11(3) - 6

➝ P(3) = 27 - 6(9) + 33 - 6

➝ P(3) = 27 - 54 + 33 - 6

➝ P(3) = 0

∴ (x-3) is a root of the Equation x³-6x²+11x-6

Then we shall divide the Equation by x-3 [Has been attached]

➝ {x³-6x²+11x-6}/x-3

➝ x²-3x+2

✭ Splitting the middle term

➝ x²-3x+2 = 0

➝ x²-2x-x+2 = 0

➝ x(x-2) - 1(x-2) = 0

➝ (x-1)(x-2) = 0

∴ The roots of the Equation are (x-1) , (x-2) & (x-3)

Answer 2

To solve:

x³ - 6x² + 11x - 6 = 0

Calculation:

Applying x = 1 by trial and error:

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

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

$\implies f(1) = 1 - 6 + 11- 6$

$\implies f(1) = 12 - 12$

$\implies f(1) = 0$

So, x = 1 is a root of the polynomial.

Now, since max power of x is 3 , we shall write (x-1) three times on the 3rd line and try to adjust the equation on the 2nd line:

$\therefore \: {x}^{3} - 6{x}^{2} + 11x - 6 = 0$

$\implies \: {x}^{3} - {x}^{2} - 5 {x}^{2} + 5x + 6x - 6 = 0$

$\implies \: {x}^{2} (x - 1) - 5x(x - 1) + 6(x - 1) = 0$

$\implies \: ({x}^{2} - 5x+ 6)(x - 1) = 0$

$\implies \: \{{x}^{2} - 3x - 2x+ 6 \}(x - 1) = 0$

$\implies \: \{x(x - 3) - 2(x - 3) \}(x - 1) = 0$

$\implies \: \{(x- 2)(x - 3) \}(x - 1) = 0$

$\implies \: (x - 1) (x- 2)(x - 3)= 0$

So, x = 1, 2 and 3