A polynomial p(x) = x⁴-2x³+3x²-ax+b, when divided by x-1 gives rem

1.	A polynomial p(x) = x⁴-2x³+3x²-ax+b, when divided by x-1 gives remainder as 5, when divided by x+1 gives remainder 19. Will this polynomial p(x) be multiple ofx-2? Support your answer with suitable working.
Answer» answer: hope it works explanation : 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

A polynomial p(x) = x⁴-2x³+3x²-ax+b, when divided by x-1 gives remainder as 5, when divided by x+1 gives remainder 19. Will this polynomial p(x) be multiple ofx-2? Support your answer with suitable working.

Answer»

answer: hope it works

explanation : 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

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A polynomial p(x) = x⁴-2x³+3x²-ax+b, when divided by x-1 gives remainder as 5, when divided by x+1 gives remainder 19. Will this polynomial p(x) be multiple ofx-2? Support your answer with suitable working.

answer: hope it works

explanation : 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

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