Question

Factorize x^3 - 5x^2 - 2x + 24

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Answer 1

Given :

• x³ - 5x² - 2x + 24

To find :

• Factorise x³ - 5x² - 2x + 24

According to the question :

Let p ( x ) = x³ - 5x² - 2x + 24

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Assuming x = 1 :

⟹ x³ = 1 × 1 × 1 = 1

⟹ 5x² = 5 × 1 × 1 = 5

⟹ 2x = 2 × 1 = 2

⟹ 24 = 24

↦p ( 1 ) = 1 - 5 - 2 + 24 = 18 ≠ 0

↦( x - 1 ) is not a factor

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Assuming x = -1 :

⟹ x³ = -1 × -1 × -1 = -1

➳ [ ( - ) × ( - ) = ( + ) , ( + ) × ( - ) = ( - ) ]

⟹ 5x² = 5 × -1 × -1 = -5

➳ [ ( + ) × ( - ) = ( - ) , ( - ) × ( + ) = ( - ) ]

⟹ 2x = 2 × -1 = 2

⟹ 24 = 24

↦p ( -1 ) = -1 -5 + 2 + 24 = 20 ≠ 0

↦( x + 1 ) is not a factor

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So, we have to search for different values of ' x ' by trial and error method.

Assuming x = 2 :

⟹ x³ = 2 × 2 × 2 = 8

⟹ 5x² = 5 × 2 × 2 = 10 × 2 = 20

⟹ 2x = 2 × 2 = 4

⟹ 24 = 24

↦p ( 2 ) = 8 - 20 - 4 + 24 = 8 ≠ 0

↦Hence, ( x - 2 ) is too not a factor

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Assuming x = -2 :

⟹ x³ = -2 × -2 × -2 = -8

➳ [ ( - ) × ( - ) = ( + ) , ( + ) × ( - ) = ( - ) ]

⟹ 5x² = 5 × -2 × -2 = -10 × 2 = -20

➳ [ ( + ) × ( - ) = ( - ) , ( - ) × ( + ) = ( - ) ]

⟹ 2x = 2 × -2 = 4

⟹ 24 = 24

↦p ( -2 ) = -8 -20 + 4 + 24 = 0

↦p ( -2 ) = 0

↦Hence, ( x + 2 ) is a Factor

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| 1 -5 -2 24

| 0 -2 +14 -24

|______________

| 1 -7 12 | 0 ( remainder )

| 0 3 -12 |___________

|______________

1 -4 | 0 ( remainder )

|_______

Thus, ( x + 2 ) ( x - 3 ) ( x - 4 ) are the factors.

∴ x³ - 5x² - 2x + 24 = ( x + 2 ) ( x - 3 ) ( x - 4 )

So, It's Done !!

Answer 2

Answer:

(x-4) (x-3) (x-2)

Step-by-step explanation:

$how \: to \: solve \: your \: question \\ \\ x {}^{3} - 5x {}^{2} - 2x + 24 \\ \\ \\ 1.find \: one \: factor \: \\ \\ factor \: by \: grouping \: \\ \\ x3 - 5x {}^{2} \: - 2x + 24 \\ x3 - 1x {}^{2} - 6x - 4x {}^{2} + 4x + 24 \\ \\ factor \: by \: grouping \: \\ \\ x {}^{3} - 1x {}^{2} - 6x - 4x {}^{2} + 4x + 24 \\ x {}^{3} - 1.x \: x \: - 6x - 4x {}^{2} + 4x + 24 \\ \\ factor \: by \: grouping \: \\ \\ x {}^{3} - 1.x \: x \: - 6x - 4x {}^{2} + 4x + 24 \\ x(x {}^{2} - x - 6) - 4(x {}^{2} - x - 6) \\ \\ factor \: by \: grouping \: \\ \\ x(x {}^{2} - x - 6) - 4(x {}^{2} - x - 6 \\ (x - 4)(x {}^{2} - x - 6) \\ \\ 2.find \: one \: factor \: \\ \\ factor \: by \: grouping \: \\ \\ (x - 4)(x {}^{2} - x - 6) \\ (x - 4)(x {}^{2} -3x + 2x - 6) \\ \\ factor \: by \: grouping \: \\ \\ (x - 4)( {x}^{2} - 3x + 2x - 6) \\ \\ factor \: by \: grouping \: \\ \\ (x - 4)(x(x - 3) + 2(x - 3)) \\ (x - 4)(x - 3)(x - 2) \\ (x - 4)(x - 3)(x - 2) \\ \\ solution..(x - 4)(x - 3)(x - 2)$

thanks post this types question

good luck ✅✅