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When we have a polynomial as 16 $x^{4}$ + 12 $x^{3}$ – 10 $x^{2}$ + 8x + 20 which is divided by 4x − 3, the quotient and the remainder are, respectively.

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x^{3}

+ 6

x^{2}

+ 2x and

\frac{61}{2}

x^{3}

+ 6

x^{2}

\frac{7}{2}

and

\frac{61}{2}

x^{2}

+ 2x +

\frac{2}{7}

and

\frac{61}{2}

x^{3}

+ 6

x^{2}

+ 2x +

\frac{7}{2}

and

\frac{61}{2}

answer is D.

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Detailed Solution

p(x) = 16

x^{4}

+ 12

x^{3}

– 10

x^{2}

+ 8x + 20 and the divisor polynomial d(x) = 4x – 3. The degree of p(x) is 4 and the degree of d(x) is 1. We are going to use the long division method. We compare the first term in p(x) and first term in d(x). We can only reach 16

x^{4}

by only multiplying

\frac{16 x^{4}}{4 x}

= 4

x^{3}

to d(x) = 4x – 3. We initiate the long division method and have,

4 x^{3} + 6 x^{2}

2 x

\frac{7}{2}

4x – 3 )

\underset{̲}{16 x^{4} + 12 x^{3} - 10 x^{2} + 8 x + 12}

− (16

x^{4}

− 12

x^{3}

)

\underset{̲}{24 x^{3} 10 x^{2}} + 8 x + 20

- (

24 x^{3} - 18 x^{2}

)

\underset{̲}{8 x^{2} + 8 x} + 20

− (8

x^{2}

− 6x )

\underset{̲}{14 x + 20}

−

(14 x - \frac{21}{2})

\underset{̲}{\frac{61}{2}}

So the remainder is

\frac{61}{2}

which is polynomial of degree 0 less that the degree of d(x). So we stop the division here find the quotient and remainder respectively as
q(x) = 4

x^{3}

+ 6

x^{2}

+ 2x +

\frac{7}{2}

and r(x) =

\frac{61}{2}

So the correct option is 4.

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When we have a polynomial as 16x4 + 12x3 – 10x2 + 8x + 20 which is divided by 4x − 3, the quotient and the remainder are, respectively.

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When we have a polynomial as 16 $x^{4}$ + 12 $x^{3}$ – 10 $x^{2}$ + 8x + 20 which is divided by 4x − 3, the quotient and the remainder are, respectively.