Question

show that (x-2) and (x-3) are factor of p(x)=x³-3x²-10x+24​

Answer 1

$\sf{Answer}$

Step by step explanation:-

Correct question:-

show that (x-2) and (x+ 3) (x -4) are factors of p(x)=x³-3x²-10x+24

We have to show that (x-2) & (x-3) are factors for

x³ -3x² -10x +24

If we substuite The given factors of cubic equation it should be equal to 0

So,

x - 2 = 0

x = 2

Substuite x =2 in Given cubic equation

x³ - 3x² -10x + 24

(2)³ - 3(2)² -10 (2) + 24

8 - 12 -20 +24

24 + 8 - 32

32 -32

0

Hence (x-2) is a factor

___________________

Now,

x + 3 = 0

x = -3

Substuite x = -3 in given cubic equation

x³ - 3x²-10x+24

(-3)³ - 3(-3)² -10(-3) +24

-27 -27 +30+24

-54 +54

0

Hence (x +3) is a factor

___________________

(x -4 ) =0

x -4 =0

x = 4

Substuite x =4 in given cubic equation

x³ -3x² -10x +24

(4)³ - 3 (4)² -10(4) +24

64-48-40+24

88-88

0

Hence (x-4) is a factor

___________________

So, we can say that (x-2) &(x+3)&(x-4) are factors of given cubic equation

This method is known as factor theoram

We can proved by factor theoram

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