Math, asked by Yash3122005, 10 months ago

Let a(x)=x^5+2x^4-x^3+2a(x)=x
5
+2x
4
−x
3
+2a, left parenthesis, x, right parenthesis, equals, x, start superscript, 5, end superscript, plus, 2, x, start superscript, 4, end superscript, minus, x, cubed, plus, 2, and b(x)=x^3+3b(x)=x
3
+3b, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 3.
When dividing aaa by bbb, we can find the unique quotient polynomial qqq and remainder polynomial rrr that satisfy the following equation:
\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}
b(x)
a(x)

=q(x)+
b(x)
r(x)

start fraction, a, left parenthesis, x, right parenthesis, divided by, b, left parenthesis, x, right parenthesis, end fraction, equals, q, left parenthesis, x, right parenthesis, plus, start fraction, r, left parenthesis, x, right parenthesis, divided by, b, left parenthesis, x, right parenthesis, end fraction,
where the degree of r(x)r(x)r, left parenthesis, x, right parenthesis is less than the degree of b(x)b(x)b, left parenthesis, x, right parenthesis.
What is the quotient, q(x)q(x)q, left parenthesis, x, right parenthesis?
q(x)=q(x)=q, left parenthesis, x, right parenthesis, equals

What is the remainder, r(x)r(x)r, left parenthesis, x, right parenthesis?
r(x)=r(x)=r, left parenthesis, x, right parenthesis, equals

Answers

Answered by amitnrw
6

Given:   a(x)  = x⁵ + 2x⁴ - x³  + 2  , b(x) = x³ + 3   a(x) = b(x)q(x) + r(x)

To find : q(x) & r(x)

Solution:

a(x)  = x⁵ + 2x⁴ - x³  + 2

b(x) = x³ + 3

                 x²  + 2x - 1

x³ + 3   _|   x⁵ + 2x⁴ - x³  + 2  |_

                  x⁵                 + 3x²

                 _______________

                          2x⁴ - x³  - 3x²  + 2

                          2x⁴                   + 6x

                         __________________

                                 - x³  - 3x² - 6x  + 2

                                 -x³                    -3

                                 ________________

                                          -3x² - 6x + 5

 

a(x)  = b(x)q(x)  +  r(x)

x⁵ + 2x⁴ - x³  + 2 = ( x³ + 3)  (x²  + 2x - 1)  + (  -3x² - 6x + 5)

q(x) = x²  + 2x - 1

r(x) =  -3x² - 6x + 5

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Answered by lhamuashoktamang
2

Answer:

Given:   a(x)  = x⁵ + 2x⁴ - x³  + 2  , b(x) = x³ + 3   a(x) = b(x)q(x) + r(x)

To find : q(x) & r(x)

Solution:

a(x)  = x⁵ + 2x⁴ - x³  + 2

b(x) = x³ + 3

                x²  + 2x - 1

x³ + 3   _|   x⁵ + 2x⁴ - x³  + 2  |_

                 x⁵                 + 3x²

                _______________

                         2x⁴ - x³  - 3x²  + 2

                         2x⁴                   + 6x

                        __________________

                                - x³  - 3x² - 6x  + 2

                                -x³                    -3

                                ________________

                                         -3x² - 6x + 5

 

a(x)  = b(x)q(x)  +  r(x)

x⁵ + 2x⁴ - x³  + 2 = ( x³ + 3)  (x²  + 2x - 1)  + (  -3x² - 6x + 5)

q(x) = x²  + 2x - 1

r(x) =  -3x² - 6x + 5

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Divide the polynomial 6x³ + 8x² by the monomial 2x. What is the ...

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On dividing a polynomial p(x) by x2 - 4, quotient and remainder ...

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Divide the polynomial (6x³ + 11x² - 10x - 7) by the binomial (2x + 1 ...

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Step-by-step explanation:

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