# Without Actual Division Prove That X4 + 2X3 + 2X2 + 2X - 3 Is Exactly Divisible By X2 + 2X - 3.

Given:

$$x^{4} + 2x^{3} – 2x^{2} – 3$$ is exactly divisible by $$x^{2} + 2x – 3$$ $$x^{2} + 2x – 3$$

=$$x^{2} + 3x – x – 3$$

=$$x (x+3) -1 (x+3)$$

= (x+3) (x-1)

Let p(x) = $$x^{4} + 2x^{3} – 2x^{2} – 3$$

=$$p(-3) = (-3)^{4} + 2(-3)^{3} – 2(-3)^{2} – 3$$

= 81 – 54 – 18 – 6 – 3

= 0

Therefore, by the converse of factor theorem, (x+3)is a factor pf p(x)

Let p(1) =$$(1)^{4} + 2(1)^{3} – 2(1)^{2} + 2(1) – 3$$

= 1 + 8 – 4 + 2 – 3

= 9 – 6 – 3

= 0

Hence by converse of factor theorem, (x-1) is a factor of p(x)

Therefore, $$x^{2} + 2x -3$$ is a factor of p(x) and p(x)is exactly divisible by $$x^{2} + 2x -3$$

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