Question

50 POINTS

Without actual division, show that x^3-3x^2-13x+15 is exactly divisible by x^2+2x-3

Answer 1

Here's your answer :-

{x}^{2} + 2x - 3 \\ = {x}^{2} + 3x - x - 3 \\ = x(x + 3) - 1(x + 3) \\ = (x + 3)(x - 1)

x + 3 = 0 \\ = > x = - 3

( { - 3})^{3} - 3 \times( { - 3})^{2} - 13 \times - 3 + 15 \\ = > - 27 - 3 \times 9 + 39 + 15 \\ = > - 27 - 27 + 54 \\ = > - 54 + 54 \\ = > 0

x - 1 = 0 \\ = > x = 1

{1}^{3} - 3 \times {1}^{2} - 13 \times 1 + 15 \\ = > 1 - 3 - 13 + 15 \\ = > - 16 + 16 \\ = > 0

as \: x + 3 \: and \: x - 1 \: is \: divisible \: by \: {x}^{3} - 3 {x}^{2} - 13x + 15 \: \\ so \: {x}^{2} + 2x - 3 \: is \: also \: divisible

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Answer 2

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,

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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,

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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.

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

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