Question

Using factor theorem factorise:-
x⁴+2x³-13x²-14x+24​

Answer 1

$\huge{\mathfrak{\underline{\red{Answer!}}}}$

(x-1) (x+2) (x-3) (x+4)

Step-by-step explanation:

Firstly, we will find the root of p(x) = x^4 + 2x³ - 13x² - 14x + 24

When x = 1

=> x^4 + 2x³ - 13x² - 14x + 24

=> (1)^4 + 2(1)³ - 13(1)² - 14(1) + 24

=> 1 + 2 - 13 - 14 + 24

=> 27 - 27

=> 0

So by factor theorem, (x - 1) is a factor

On dividing p(x) by (x - 1), we get x³ + 3x² - 10x -24

☆ Refer to the attachment 1 ☆

Now find the root of p(x) = x³ + 3x² - 10x -24

When x = -2

=> x³ + 3x² - 10x -24

=> (-2)³ + 3(-2)²- 10(-2) -24

=> -8 + 12 + 20 - 24

=> -32 + 32

=> 0

So by factor theorem, (x + 2) is a factor

On dividing p(x) by (x + 2), we get x² + x - 12

☆ Refer to the attachment 2 ☆

Now, Using middle term splitting :

=> x² + x - 12

=> x² + 4x - 3x - 12

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

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

∴ x^4 + 2x³ - 13x² - 14x + 24 = (x-1) (x+2) (x-3) (x+4)

Answer 2

Answer:

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