Question

factorise x^3-6x+4.​

Answer 1

x^3 – 6x + 4 = 0

Let us consider the expression, x^3 -6x +4

We are going to use the trial and error method to factorise this expression,

Let us consider the value of x to e 1.

When x = 1

(1)^3 – 6(1) + 4

= 1 – 6 + 4

= -1

Therefore, x = 1 is not the required value.

When x = 2

(2)^3 – 6(2) + 4

= 8 -12 + 4

= 12 – 12

= 0

Therefore, the value of the expression becomes 0 when the value of x is equal to 2.

Hence, (x-2) is a factor.

Now,

X^3 – 6x + 4

= x^3 – 2x^2 +2x^2 – 4x -2x +4

= x^2(x-2) +2x(x-2) -2(x-2)

= (x-2)(x^2 +2x -2)

Therefore the given equation can also be written as,

(x-2)(x^2 + 2x - 2) = 0

Since, the product of two expressions is 0, then the value of either of the expression must be 0.

Therefore, x-2 = 0

x = 2

Again, x^2 +2x – 2 = 0

Using Sridharacharya’s rule for solving quadratic equation we get,

x = -1 + i

x = -1 – i

Therefore,

x = 2, -1 + i, -1 -i

Answer 2

Answer:

Step-by-step explanation:

$x^3 – 6x + 4 = 0$

Let us consider the expression, $x^3 -6x +4$

We are going to use the trial and error method to factorise this expression,

Let us consider the value of x to e 1.

When x = 1

$(1)^3 – 6(1) + 4$

$= 1 – 6 + 4$

=$-1$

Therefore, x = 1 is not the required value.

When $x = 2$

$(2)^3 – 6(2) + 4$

= $8 -12 + 4$

= $12 – 12$

= $0$

Therefore, the value of the expression becomes 0 when the value of x is equal to 2.

Hence, (x-2) is a factor.

Now,

$X^3 – 6x + 4$

= $x^3 – 2x^2 +2x^2 – 4x -2x +4$

=$x^2(x-2) +2x(x-2) -2(x-2)$

= $(x-2)(x^2 +2x -2)$

Therefore the given equation can also be written as,

$(x-2)(x^2 + 2x - 2) = 0$

Since, the product of two expressions is 0, then the value of either of the expression must be 0.

Therefore,$x-2 = 0$

$x = 2$

Again,$x^2 +2x – 2 = 0$

Using Sridharacharya’s rule for solving quadratic equation we get,

$x = -1 + i$

$x = -1 – i$

Therefore,

$x = 2, -1 + i, -1 -i$

This is the required answer.

hope it helps

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