Math, asked by Rehan0917, 1 year ago

without actual division prove that x^3 - 3 x^2 - 13 x + 15 is exactly divisible by X ^2 + 2 x - 3

Answers

Answered by abhi569
288
x^2 + 2x - 3 = x^2 + 3x - x -3
 
                   = x(x + 3)-1(x + 3)
 
                   = (x + 3)(x  - 1)

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checking whether (x + 3) and (x - 1) are the factors or not,

==========================

Taking x + 3 = 0    So, x = -3 

Putting the value of x in given equation,

x^3 - 3x^2 - 13x + 15 = 0 

(-3)^3 - 3(-3)^2 - 13(-3) + 15 = 0 

-27 - 3(9) + 39 + 15 = 0 

-27 -27 + 39  + 15 = 0

-54 + 54 = 0 

0 =0 

Hence, (x + 3) is the factor,

=======================


checking for (x -1)  x  = 1 

Putting the value of x in given equation,

x^3 - 3x^2 - 13x + 15 = 0 

(1)^3 - 3(1)^2 - 13(1) + 15 =0 

1 - 3 - 13 + 15 = 0 

-15 + 15 = 0 

0 =0 

Hence, (x - 1) is also a factor.



=========================

Then, x^2 + 2x - 3 is the factor of given equation



i hope this will help you


(-:
Answered by adithrnair
3

Answer:

x^2 + 2x - 3 is the factor of given equation

Step-by-step explanation:

x^2 + 2x - 3 = x^2 + 3x - x -3

 

                  = x(x + 3)-1(x + 3)

 

                  = (x + 3)(x  - 1)

===============================

checking whether (x + 3) and (x - 1) are the factors or not,

==========================

Taking x + 3 = 0    So, x = -3  

Putting the value of x in given equation,

x^3 - 3x^2 - 13x + 15 = 0  

(-3)^3 - 3(-3)^2 - 13(-3) + 15 = 0  

-27 - 3(9) + 39 + 15 = 0  

-27 -27 + 39  + 15 = 0

-54 + 54 = 0  

0 =0  

Hence, (x + 3) is the factor,

=======================

checking for (x -1)  x  = 1  

Putting the value of x in given equation,

x^3 - 3x^2 - 13x + 15 = 0  

(1)^3 - 3(1)^2 - 13(1) + 15 =0  

1 - 3 - 13 + 15 = 0  

-15 + 15 = 0  

0 =0  

Hence, (x - 1) is also a factor.

=========================

Then, x^2 + 2x - 3 is the factor of given equation

i hope this will help you

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