Question

Divide 3x4 -5x3 + 4x2 + 3x – 5 by x2 -3, and verify the division algorithm.

Answer 1

we have to divide 3x⁴ - 5x³ + 4x² + 3x - 5 by x² - 3

x² - 3 ) 3x⁴ - 5x³ + 4x² + 3x - 5 (3x² - 5x + 13

3x⁴ - 9x²

.....................................

-5x³ + 13x² + 3x

-5x³ + 15x

.....................................................

13x² -12x - 5

13x² - 39

.................................................

-12x + 34

therefore, remainder = -12x + 34

and quotient = 3x² - 5x + 13

verification : using division algorithm, a = bq + r, 0 ≤ r < b

here, a = 3x⁴ - 5x³ + 4x² + 3x - 5

b = x² - 3

q = 3x² - 5x + 13

r = 18x + 34

LHS = 3x⁴ - 5x³ + 4x² + 3x - 5

RHS = bq + r

= (x² - 3)(3x² - 5x + 13) + (-12x + 34)

= 3x⁴ - 5x³ + 13x² - 9x² + 15x - 39 - 12x + 34

= 3x⁴ - 5x³ + 4x² + 3x - 5

here LHS = RHS hence verified

Answer 2

$\huge\mathcal{Answer:}$

${x}^{2} \sqrt{3 {x}^{4} } - 5 {x}^{3} + 4 {x}^{2} + 3x - 5|3 {x}^{2} - 5x + 13$

$3 {x}^{4} - 9 {x}^{2}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀

$- 5 {x}^{3} + 13 {x}^{2} + 3x$

$- 5 {x}^{3} + 15x$

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$13 {x}^{2} - 12x - 5$

$13 {x}^{2} - 39$

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$- 12x + 34$

$\huge\mathcal{Verification:}$

$a = bq + r \: o \leqslant r \leqslant b$

$a = 3 {x}^{4} - 5 {x}^{3} + 4 {x}^{2} + 3x - 5$

$b = {x}^{2} - 3$

$q = 3{x}^{2} - 5x + 13$

$r = 18x + 34$

$lhs = 3 {x}^{4} - 5 {x}^{3} + 4 {x}^{2} + 3x - 5$

$rhs = bq + r$

$= ( {x}^{2} - 3)(3 {x}^{2} - 5x + 13) + ( - 12x + 34)$

$= 3 {x}^{4} - 5 {x}^{3} + 13 {x}^{2} - 9 {x}^{2} + 15x - 39 - 12x + 34$

$3 {x}^{2} - 5 {x}^{3} + 4 {x}^{2} + 3x - 5$

$lhs = rhs$