what must be added to the polynomial $f (x) = 2 x^{3} - 2 x^{2} + x - 1$ so that the resulting polynomial is exactly divisible by $x^{2} + 2 x - 3$ ?

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Solution

Dividing $2 x^{3} - 2 x^{2} + x - 1$ from $x^{2} + 2 x - 3$ we get

$\frac{2 x^{3} - 2 x^{2} + x - 1}{x^{2} + 2 x - 3}$ =

$\begin{matrix} 2 x - 6 - ----- - x^{2} + 2 x - 3 | 2 x^{3} - 2 x^{2} + x - 1 - (2 x^{3} + 4 x^{2} - 6 x) - --------------------- - - 6 x^{2} + 7 x - 1 - (- 6 x^{2} - 12 x + 18) - ------------------------ - 19 x - 19 \end{matrix}$

Since the remainder is $19 x - 19$ we can subtract it or add $- (19 x - 19)$ to the given polynomial to make it divisible by $x^{2} + 2 x - 3$

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Similar questions

f (x) = 4 x^{4} + 2 x^{3} - 2 x^{2} + x - 1

g (x) = x^{2} + 2 x - 3

The expression that should be added to the polynomial f(x) = x⁴ + 2x³ – 2x² + x + 1, so that it should be exactly divisible by (x² + 2x – 3) is

p (x) = 2 x^{3} + 9 x^{2} - 5 x - 15

2 x + 3

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