Question

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Using remainder therom, find the remainder when p(x) is divided by g(x) :-

1) p(x) = x⁴ - 3x² + 4x - 12 g(x) = x - 3

2) p(x) = 4x³ - 12x² + 11x - 3 g(x) = x + ½

3) p(x) = 3x⁴ + 2x³ - x²/3 - x/9 + 2/27 g(x) = x + ⅔

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Answer 1

Questions

Using the remainder theorem, find the remainder when p(x) is divided by g(x) -:

(i) p(x) = x⁴ - 3x² + 4x - 12 g(x) ; x - 3

(ii) p(x) = 4x³ - 12x² + 11x - 3 g(x) ; x + ½

(iii) p(x) = 3x⁴ + 2x³ - x²/3 - x/9 + 2/27 g(x) ; x + ⅔

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Answers

(i) Let's find the value of 'x' in g(x).

⇒ x - 3 = 0

⇒ x = 3

Now, using the remainder theorem we will divide p(x) by g(x) where p(3).

⇒ p(3) = (3)⁴ - 3(3)² + 4(3) - 12

⇒ p(3) = 81 - 3 × 9 + 12 - 12

⇒ p(3) = 81 - 27 + 0

⇒ p(3) = 54

∴ The remainder is 54 when p(x) is divided by g(x) in the given question.

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(ii) Let's find the value of 'x' in g(x).

⇒ x + ¹/₂

⇒ x = ⁻¹/₂

Now, using the remainder theorem we will divide p(x) by g(x) where p(⁻¹/₂).

$\sf \implies p(\dfrac{-1}{2} ) = 4(\dfrac{-1}{2})^{3} - 12(\dfrac{-1}{2})^{2} + 11 (\dfrac{-1}{2}) - 3$

$\sf \implies p(\dfrac{-1}{2}) = 4 \times \dfrac{-1}{8} - 12 \times \dfrac{1}{4}+11 \times \dfrac{-1}{2} -3$

$\sf \implies p(\dfrac{-1}{2}) = \dfrac{-4}{8} - \dfrac{12}{4}+\dfrac{-11}{2} -3$

$\sf \implies p(\dfrac{-1}{2}) = \dfrac{-1}{2} - 3+\dfrac{-11}{2} -3$

$\sf \implies p(\dfrac{-1}{2}) = \dfrac{-1}{2} +\dfrac{-11}{2} -3- 3$

$\sf \implies p(\dfrac{-1}{2}) = \dfrac{-12}{2}-6$

$\sf \implies p(\dfrac{-1}{2}) = -6-6$

$\sf \implies p(\dfrac{-1}{2}) = -12$

∴ The remainder is (-12) when p(x) is divided by g(x) in the given question.

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(iii) Let's find the value of 'x' in g(x).

⇒ x + ⅔ = 0

⇒ x = ⁻⅔

Now, using the remainder theorem we will divide p(x) by g(x) where p(⁻⅔).

$\sf \implies p(\dfrac{-2}{3}) = 3(\dfrac{-2}{3})^{4}+2(\dfrac{-2}{3})^{3} - \dfrac{(\frac{-2}{3} )^{2}}{3} - \dfrac{\frac{-2}{3}}{9} + \dfrac{2}{27}$

$\sf \implies p(\dfrac{-2}{3}) = 3\times \dfrac{16}{81}+2\times \dfrac{-8}{27} - \dfrac{\frac{4}{9}}{3} - \dfrac{\frac{-2}{3}}{9} + \dfrac{2}{27}$

$\sf \implies p(\dfrac{-2}{3}) = \dfrac{48}{81}+ \dfrac{-16}{27} - \dfrac{4}{27} - \dfrac{-2}{27} + \dfrac{2}{27}$

$\sf \implies p(\dfrac{-2}{3}) = \dfrac{48}{81}+ \dfrac{-16}{27}$

$\sf \implies p(\dfrac{-2}{3}) = \dfrac{48}{81}+ \dfrac{-48}{81}$

$\sf \implies p(\dfrac{-2}{3}) = \dfrac{0}{81}$

$\sf \implies p(\dfrac{-2}{3}) =0$

∴ The remainder is 0 when p(x) is divided by g(x) in the given question.

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Answer 2

1) p(x) = x⁴ - 3x² + 4x - 12 g(x) = x - 3

Here,

• x - 3 = 0
• x = 3

Now, putting the value of x in the polynomial,

$\implies\:p(x) = x^4 - 3x^2+ 4x - 12$

$\implies\:p(3) = 3^4 - 3(3)^2 + 4(3) - 12$

$\implies\:p(3) = 81 - 27 + 12 - 12$

${\boxed{\implies\:p(3) = 54}}$

Hence, the remainder of p(x) = x⁴ - 3x² + 4x - 12 is 54.

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2) p(x) = 4x³ - 12x² + 11x - 3 g(x) = x + ½

Here,

• x + ½ = 0
• x = - ½

Now, putting the value of x in polynomial,

$\implies\:p(x) = 4x^3 - 12x^2 + 11x - 3$

$\sf \implies\:p(\dfrac{-1}{2}) = 4(\dfrac{-1}{2})^3- 12(\dfrac{-1}{2})^2+ 11(\dfrac{-1}{2}) - 3$

$\sf \implies\:p(\dfrac{-1}{2}) = 4\times\dfrac{-1}{8}- 12\times\dfrac{1}{4}+ 11\times\dfrac{-1}{2} - 3$

$\sf\implies\:p(\dfrac{-1}{2}) = \dfrac{-1}{2}- 3+ \dfrac{-11}{2}- 3$

$\implies\:p(\dfrac{-1}{2}) = \dfrac{-1-11}{2}-3-3$

$\implies\:p(\dfrac{-1}{2}) = \dfrac{-12}{2}-6$

$\implies\:p(\dfrac{-1}{2}) = -6-6$

${\boxed{\implies\:p(\dfrac{-1}{2}) = -12}}$

Hence, The Remainder of p(x) = 4x³ - 12x² + 11x - 3 is -12.

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3) p(x) = 3x⁴ + 2x³ - x²/3 - x/9 + 2/27 g(x) = x + ⅔

Here,

• x + ⅔ = 0
• x = -⅔

Now, Putting the value of x in polynomial,

$\implies\:p(x) = 3x^4 + 2x^3 - \dfrac{x^2}{3} - \dfrac{x}{9} + \dfrac{2}{27}$

$\sf \implies p(\dfrac{-2}{3}) = 3(\dfrac{-2}{3})^{4}+2(\dfrac{-2}{3})^{3} - \dfrac{(\frac{-2}{3} )^{2}}{3} - \dfrac{\frac{-2}{3}}{9} + \dfrac{2}{27}$

$\sf \implies p(\dfrac{-2}{3}) = 3\times \dfrac{16}{81}+2\times \dfrac{-8}{27} - \dfrac{\frac{4}{9}}{3} - \dfrac{\frac{-2}{3}}{9} + \dfrac{2}{27}$

$\sf \implies p(\dfrac{-2}{3}) = \dfrac{48}{81}+ \dfrac{-16}{27} - \dfrac{4}{27} - \dfrac{-2}{27} + \dfrac{2}{27}$

$\implies p(\dfrac{-2}{3}) = \dfrac{48}{81}+ \dfrac{-16}{27} - \dfrac{4}{27}+ \dfrac{4}{27}$

$\implies p(\dfrac{-2}{3}) = \dfrac{48}{81}+ \dfrac{-16}{27}$

$\implies p(\dfrac{-2}{3}) = \dfrac{48-48}{81}$

${\boxed{\implies p(\dfrac{-2}{3}) =0}}$

Hence, The Remainder of p(x) = 3x⁴ + 2x³ - x²/3 - x/9 + 2/27 is 0.

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