Question

FI
25. Factorize : x3 = 3x2 – 10 x + 24​

Answer 1

Answer:

Factors of x³-3x²-10x+24

Factors of x³-3x²-10x+24 = (x-2)(x-4)(x+3)

Step-by-step explanation:

x³-3x²-10x+24

= x³+(-2x²-x²)+(2x-12x)+24

=(x³-2x²)+(-x²+2x)+(-12x+24)

=x²(x-2) -x(x-2) -12(x-2)

=(x-2)(x²-x-12)

=(x-2)(x²-4x+3x-12)

=(x-2)[x(x-4)+3(x-4)]

=(x-2)[(x-4)(x+3)]

=(x-2)(x-4)(x+3)

Therefore,

Factors of x³-3x²-10x+24

= (x-2)(x-4)(x+3)

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Answer 2

Correct Question:

Factorize : x³ -3x² -10x +24

Solution:

Let p(x) = x³ -3x² -10x +24

Using hit and trial method

Taking x = 1

$\text{p(1) = } \: {(1)}^{3} - 3 {(1)}^{2} - 10(1) + 24 \\ \\ \text{p(1) = }1 - 3 - 10 + 24 \\ \\ \text{p(1) = 25 - 13} \\ \\ \longrightarrow 12 \ne0$

Taking x = 2

$\text {p(2) = } {(2)}^{3} - 3 {(2)}^{2} - 10(2) + 24 \\ \\ \text{p(2) = }8 - 3(4) - 20 + 24 \\ \\ \text{p(2) = 8 - 12 - 20 + 24} \\ \\ \text{p(2) = 32 - 32} \\ \\ \longrightarrow \: 0 = 0$

Therefore, (x-2) is a factor of p(x)

Dividing p(x) by (x-2)

(Refer to the attachment)

$\implies \: {x}^{3} - 3 {x}^{2} - 10x + 24 \\ = (x - 2)( {x}^{2} - x - 12)$

Splitting the middle term

= (x-2) [x²-4x +3x -12]

= (x-2) [ x(x-4) +3(x-4) ]

= (x-2) [ (x+3) (x-4) ]

= (x-2) (x+3) (x-4)

Thus, x³ -3x² -10x +24 is factorized as (x-2)(x+3)(x-4)