Question

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Answer 1

When f(x) is divided by x-1 and x+1 the remainder are 5 and 19 respectively.

∴f(1)=5 and f(−1)=19

⇒(1)

4

−2×(1)

3

+3×(1)

2

−a×1+b=5

and (−1)

4

−2×(−1)

3

+3×(−1)

2

−a×(−1)+b=19

⇒1−2+3−a+b=5

and 1+2+3+a+b=19

⇒2−a+b=5 and 6+a+b=19

⇒−a+b=3 and a+b=13

Adding these two equations, we get

(−a+b)+(a+b)=3+13

⇒2b=16⇒b=8

Putting b=8 and −a+b=3, we get

−a+8=3⇒a=−5⇒a=5

Putting the values of a and b in

f(x)=x

4

−2x

3

+3x

2

−5x+8

The remainder when f(x) is divided by (x-2) is equal to f(2).

So, Remainder =f(2)=(2)

4

−2×(2)

3

+3×(2)

2

−5×2+8=16−16+12−10+8=10

Answer 2

$\large\underline{\bold{Given \:Question - }}$

$\sf \: p(x)= {x}^{4} - {2x}^{3} + {3x}^{2} - ax + b  is \: a \: polynomial \: such \: that \\ \sf \: when \: it \: is \: divided \: by  \: x−1 \:  and \:  x+1, \: the \: remainder \: \\ \sf \: are \: respectively \:  5  \: and  \: 19, \: find \: value \: of \: a \: and \: b. \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: Determine \: the \: remainder \: when \:  p(x) \:  is \: divided \: by  \: (x−2).$

$\large\underline{\sf{Solution-}}$

Given polynomial is

$\rm :\longmapsto\:p(x) = {x}^{4} - {2x}^{3} + {3x}^{2} - ax + b$

Further given that,

When p(x) is divided by x - 1, the remainder is 5.

We know,

Remainder Theorem states that if a polynomial p(x) is divided by linear polynomial x - a, the remainder is p(a)

So, using Remainder Theorem, we have

$\rm :\longmapsto\:p(1) = 5$

$\rm :\longmapsto\:{1}^{4} - {2(1)}^{3} + {3(1)}^{2} - a(1) + b = 5$

$\rm :\longmapsto\:1- 2 + 3 - a + b = 5$

$\rm :\longmapsto\:2 - a + b = 5$

$\bf\implies \:b = 3 + a - - - - (1)$

Again, Given that

When p(x) is divided by (x + 1), the remainder is 19.

So, using Remainder Theorem, we have

$\rm :\longmapsto\:p( - 1) = 19$

$\rm :\longmapsto\:{( - 1)}^{4} - {2( - 1)}^{3} + {3( - 1)}^{2} - a( - 1) + b = 19$

$\rm :\longmapsto\:1 + 2 + 3 + a + b = 19$

$\rm :\longmapsto\:6 + a + b = 19$

$\rm :\longmapsto\:a + b = 19 - 6$

$\rm :\longmapsto\:a + b = 13$

$\rm :\longmapsto\:a + 3 + a = 13 \: \: \: \: \: \: \: \{using \: (1) \: \}$

$\rm :\longmapsto\:2a + 3 = 13 \:$

$\rm :\longmapsto\:2a = 13 - 3$

$\rm :\longmapsto\:2a = 10$

$\bf\implies \:a = 5$

Therefore,

$\bf\implies \:b = 3 + a = 3 + 5 = 8$

Hence,

$\bf :\longmapsto\:p(x) = {x}^{4} - {2x}^{3} + {3x}^{2} - 5x + 8$

Now, the remainder when p(x) is divided by x - 2 is given by

$\bf :\longmapsto\:p(2)$

$\rm \: = \:\:{2}^{4} - {2(2)}^{3} + {3(2)}^{2} - 5(2) + 8$

$\rm \: = \:16 - 16 + 12 - 10 + 8$

$\rm \: = \:20 - 10$

$\bf \: = \:10$

$\begin{gathered}\begin{gathered}\bf\: Hence-\begin{cases} &\bf{a = 5} \\ \\ &\bf{b = 8}\\ \\ &\bf{Remainder = 10} \end{cases}\end{gathered}\end{gathered}$