Question 1: Find the remainder when x^2 - 8x + 4 is divided by 2x + 1.  

Answer:

\displaystyle \text{Step 1} : 2x + 1 = 0 \Rightarrow x = \frac{-1}{2}

\displaystyle \text{Step 2:  Required remainder = Value of given polynomial } x^2 - 8x + 4 \\ \\ \text{ at } x = \frac{-1}{2}

\displaystyle \text{Therefore Remainder } = \Big(\frac{-1}{2} \Big)^2 - 8 \Big( \frac{-1}{2} \Big) + 4 = \frac{1}{4}+4+4=8\frac{1}{4}

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Question 2: Find the value of 'a' if the division of ax^3 + 9x^2 + 4x - 10 by x + 3 leaves a remainder of 5.

Answer:

x+3=0 \Rightarrow  x=-3

\text{Given, remainder is 5;  therefore: }

The value of ax^3 + 9x^2 + 4x - 10 at x =-3 is 5

\Rightarrow  a(-3)^3 + 9(-3)^2 + 4(-3) - 10 = 5

\Rightarrow -27a+ 81-12-10=5 \text{ or }  a=2

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Question 3: When the polynomial 2x^3 -kx^2 + (5k- 3)x- 8 is divided by, x- 2, the remainder is 14. Find the value of 'k'.

Answer:

x-2 = 0 \Rightarrow  x = 2

Given. remainder is 14, therefore:

\Rightarrow  2(2)^3 - k(2)^2 + ( 5k-3) \times 2 - 8 = 14

\Rightarrow  16 - 4k +10k - 6-8 = 14

\Rightarrow  6k = 12

\Rightarrow  k = 2

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Question 4: The polynomials 3x^3 - ax^2 + 5x- 13 and (a + 1)x^2 -7x+ 5 leave the same remainder when divided by x - 3. Find the value of 'a'.

Answer:

x-3 =0 \Rightarrow  x=3

Since, the given polynomials leave the same remainder when divided by x - 3

Value of polynomial 3x^3 - 9x^2 + 5x - 13 \text{ at } x = 3 is the same as the value of polynomial (a + 1) x^2 - 7x + 5 \text{ at } x = 3

\Rightarrow  3(3)^3 - 9(3)^2 + 5 \times 3 - 13  = ( a+1)(3)^2 - 7 \times 3 + 5

\Rightarrow  81 - 9a + 15 - 13 = 9a + 9 - 21 + 5

\Rightarrow  18a = 90

\Rightarrow  a = 5

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Question 5: When f(x) = x^3 + ax^2 - bx - 8 is divided by x - 2, the remainder is zero and when divided by x + 1, the remainder is -30. Find the values of a and b.

Answer:

Since, x - 2 = 0 \Rightarrow x = 2

Given that on dividing f(x) = x^3 + ax^2 - bx - 8 \text{ by }  x - 2 the remainder = 0

\Rightarrow f(2) = 0

\Rightarrow (2)^3 + a(2)^2 - b(2) - 8 = 0

\Rightarrow 8 + 4a - 2b - 8 = 0

\Rightarrow 4a - 2b = 0

\Rightarrow 2a - b = 0 \text{     ... ... ... ... ... i)}

Given that on dividing f(x) = x^3 + ax^2 - bx - 8 \text{ by }  x +1 the remainder = -30

\Rightarrow f(-1) = -30

\Rightarrow (-1)^3 + a( -1)^2 - b(-1) - 8 = -30

\Rightarrow a+b = 21 \text{     ... ... ... ... ... ii)}

Solving i) and ii) we get a = -7 and b = -14

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Question 6: What number should be added to 2x^3 - 3x^2 + x so that when the resulting polynomial is divided by x - 2, the remainder is 3 ?

Answer:

Let the number added be k so the resulting polynomial is 2x^3 - 3x^2 + x + k

Given, when this polynomial is divided by x- 2, the remainder = 3

\Rightarrow 2(2)^3 - 3 ( 2)^2 + 2 + k = 3

\Rightarrow 16-12+2+k = 3

\Rightarrow k = - 3

Therefore the required number to be added = - 3 

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Question 7: Determine whether x-1   is a factor of x^6 - x^5 + x^4 + x^3 - x^2 - x + 1 or not?

Answer:

x-1 = 0 \Rightarrow x = 1

Therefore when given polynomial is divided by x-1, the remainder

= (1)^6 - (1)^5 + (1)^4 + (1)^3 - (1)^2 - (1) + 1 = 1 - 1 + 1 +1-1-1+1 = 4-3 = 1

which is not equal to zero.

Therefore x-1 is not a factor of the given polynomial.

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Question 8: If x- 2 is a factor of x^2 - 7x + 2a, find the value of a.

Answer:

x-2 = 0 \Rightarrow x = 2

Since, x-2 is a factor of polynomial x^2 - 7x + 2a

Therefore remainder = 0

\Rightarrow (2)^2 - 7(2) + 2a =0

\Rightarrow a = 5

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Question 9: Find the value of 'k' if (x -2) is a factor of x^3 + 2x^2 - kx +10. Hence, determine whether (x + 5) is also a factor. [ICSE Board 2011]

Answer:

x - 2 is a factor and x - 2 = 0 \Rightarrow x = 2

Therefore The value of given expression x^3 + 2x^2 - kx +10 is zero at x = 2

\Rightarrow \text{ Remainder } = 0

\Rightarrow (2)^3 + 2(2)^2 - k \times 2 + 10 = 0

\Rightarrow 8 + 8 -2k + 10 = 0

\Rightarrow k = 13

On substituting k = 13, the given expression becomes x^3 + 2x^2 - 13x +10.

Now to check whether (x + 5) is also a factor or not, find the value of the given expression for x =-5

Therefore x^3 + 2x^2 - 13x +10 \text{ at } x = -5

= (-5)^3 + 2(-5)^2 - 13(-5) +10 = -125 + 50 + 65 + 10 = -125 + 125 =0

Since, the remainder is 0

\Rightarrow (x+ 5) is a factor.

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Question 10: Given that x + 2 and x - 3 are factors of x^3 + ax + b; calculate the values of a and b.

Answer:

Given, x+ 2 is a factor of x^3 + ax + b;

\Rightarrow (-2)^3 + a(-2) + b = 0

\Rightarrow -2a + b = 8 \text{     ... ... ... ... ... i)}

Again, given that : x-3 is a factor of  x^3 + ax + b;

\Rightarrow (3)^3 + a(3) + b = 0

\Rightarrow 3a+b = -27 \text{     ... ... ... ... ... ii)}

Solving i) and ii) we get a = -7 and b = -6

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Question 11: Polynomial x^3 - ax ^2 + bx - 6 leaves remainder -8 when divided by x-1 and x-2 is a factor of it. Find the values of a and b.

Answer:

On dividing by x -1 , the polynomial x^3 - bx^2 + bx - 6 leaves remainder -8.

\Rightarrow  (1)^3 - a(1)^2 + b(1) - 6 = - 8

\Rightarrow  - a + b = - 3

\Rightarrow  a - b = 3  \text{     ... ... ... ... ... i)}

(x-2) is a factor of polynomial x^3 - bx^2 + bx - 6

\Rightarrow (2)^2 - a(2)^2 + b( 2) - 6 = 0

\Rightarrow 8 - 4a + 2b - 6 = 0

\Rightarrow 2a - b = 1  \text{     ... ... ... ... ... ii)}

Solving i) and ii) we get a = -2 and b = -5

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Question 12: Using the Factor Theorem, show that (x - 2) is a factor of 3x^2-5x-2. Hence, factorize the given expression.

Answer:

Since x-2=0

\Rightarrow x=2

Therefore Remainder = The value of 3x^2-5x-2 at x = 2

\begin{array}{rll} x-2& \overline{)3x^2-5x-2(} &3x+1 \\ (-) & \underline{3x^2-6x} & \\    & {\hspace{1.0cm} x-2 }  &    \\   (-) & {\hspace{1.0cm} \underline{x-2 } } & \\ & {\hspace{1.5cm}\times } & \end{array}

= 3(2)^2 - 5(2) - 2 = 12 - 10 -2 = 0

\Rightarrow (x-2) is a factor of 3x^2 - 5x - 2

Now dividing ( 3x^2 - 5x - 2) by ( x-2), we get quotient = 3x + 1

Therefore 3x^2 - 5x - 2 = (x-2)(3x+1)

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Question 13: Show that 2x+7 is a factor of 2x^3 + 5x^2 - 11x - 14. Hence factorize the given expression completely, using the factor theorem. [ICSE Board 2006].

Answer:

\displaystyle 2x+7 = 0 \Rightarrow x = \frac{-7}{2}

\displaystyle \text{Remainder }= \text{ Value of } 2x^3 + 5x^2 - 11x - 14 \text{ at } x = \frac{-7}{2}

\displaystyle = 2\Big(\frac{-7}{2} \Big)^3 + 5\Big(\frac{-7}{2} \Big)^2 - 11\Big(\frac{-7}{2} \Big) - 14

\displaystyle = - \frac{343}{4} + \frac{245}{4} + \frac{77}{2} - 14

\displaystyle = \frac{-343+245+154-56}{4} = 0

\displaystyle \Rightarrow (2x+7) \text{ is a factor } \text{ of }  2x^3 + 5x^2 - 11x - 14

\begin{array}{rll} 2x+7& \overline{)2x^3 + 5x^2 - 11x - 14(} & x^2-x-2 \\ (-) & \underline{2x^3+7x^2} & \\    & {\hspace{1.0cm} -2x^2-11x-14 }  &    \\   (-) & {\hspace{1.0cm} \underline{-2x^2-7x } } &  \\ & {\hspace{2.0cm}-4x-14 } &  \\ (-) & {\hspace{2.0cm}\underline{-4x-14 } }&  \\ & {\hspace{2.5cm}\times } & \end{array}

\displaystyle \therefore 2x^3 + 5x^2 - 11x - 14 = (2x+7)( x^2-x-2)

\displaystyle = ( 2x+7) ( x^2 - 2x + x - 2)

\displaystyle = ( 2x+7) [x(x-2)+ 1(x-2)]

\displaystyle = ( 2x+ 7) ( x-2)(x+1)

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Question 14: Using the Remainder Theorem, factorize the expression 2x^3 + x^2 - 2x - 1 completely.

Answer:

First Step : For x = 1, the value of given expression

= 2(1)^3 + (1)^2 - 2 ( 1) - 1 = 2 + 1 - 2 - 1 = 0

\Rightarrow x-1 is a factor of 2x^3 + x^2 - 2x - 1

\begin{array}{rll} x-1& \overline{)2x^3 + x^2 - 2x - 1(} & 2x^2 + 3x + 1 \\ (-) & \underline{2x^3-2x^2} & \\    & {\hspace{1.0cm} 3x^2-3x-1 }  &    \\   (-) & {\hspace{1.0cm} \underline{3x^2-3x } } &  \\ & {\hspace{2.0cm}x-1 } &  \\ (-) & {\hspace{2.0cm}\underline{x-1 } }&  \\ & {\hspace{2.5cm}\times } & \end{array}

Second Step:

2x^3 + x^2 - 2x - 1 = (x-1)( 2x^2 + 3x + 1)

= ( x-1) ( 2x^2 + 3x + 1)

= ( x-1)(2x^2 + 2x + x + 1)

= ( x-1)[ 2x( x+1)+1(x+1)]

= ( x-1)(x+1)(2x+1)

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Question 15: Find the values of a and b so that the polynomial x^3 + ax^2 + bx - 45 has (x - 1) and (x + 5) as its factors. For the values of a and b , as obtained above, factorize the given polynomial completely.

Answer:

(x - 1) is a factor of given polynomial x^3 + ax^2 + bx - 45

\Rightarrow (1)^3 + a(1)^2 + b(1) - 45 = 0

\Rightarrow a+b = 44   \text{     ... ... ... ... ... i) }

(x +5) is a factor of given polynomial x^3 + ax^2 + bx - 45

\Rightarrow (-5)^3 + a(-5)^2 + b(-5) - 45 = 0

\Rightarrow 5a- b = 34   \text{     ... ... ... ... ... ii) }

On solving equations i) and ii), we get : a = 13 and b = 31

Therefore the given polynomial  x^3 + ax^2 + bx - 45 = x^3 + 13x^2 + 31x - 45

Now divide this polynomial by ( x-1)   we get

\begin{array}{rll} x-1& \overline{)x^3+13x^2+31x-45(} & x^2+14x+45 \\ (-) & \underline{x^3-x^2} & \\    & {\hspace{1.0cm} 14x^2+31x-45 }  &    \\   (-) & {\hspace{1.0cm} \underline{14x^2-14x } } &  \\ & {\hspace{2.0cm}45x-45 } &  \\ (-) & {\hspace{2.0cm}\underline{45x-45 } }&  \\ & {\hspace{2.5cm}\times } & \end{array}

x^3 + 13x^2 + 31x - 45 = (x-1)(x^2 + 14 x + 45)

= (x-1)( x^2 + 9x + 5x + 45)

= ( x-1)[ x(x+9) + 5 ( x + 9)]

= ( x-1)( x+9)(x+5)

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Question 16: If (x - 2) is a factor of 2x^3 - x^2 - px - 2

(i) find the value of p.

(ii) with the value of p, factorize the above expression completely

Answer:

(i) x-2=0 \Rightarrow  x=2

Since, (x - 2) is a factor of given expression \Rightarrow \text{ Remainder } = 0

\Rightarrow 2(2)^3 - (2)^2 - p \times 2 - 2 = 0

\Rightarrow 10-2p = 0

\Rightarrow p = 5

(ii) \therefore 2x^3 - x^2 - px - 2 = 2x^3 - x^2 - 5x - 2

On dividing 2x^3 - x^2 - 5x - 2 by (x-2) we get:

\begin{array}{rll} x-2& \overline{)2x^3-x^2-5x-2(} & 2x^2 + 3x + 1 \\ (-) & \underline{2x^3-4x^2} & \\    & {\hspace{1.0cm} 3x^2-5x-2 }  &    \\   (-) & {\hspace{1.0cm} \underline{3x^2-6x } } &  \\ & {\hspace{2.0cm}x-2 } &  \\ (-) & {\hspace{2.0cm}\underline{x-2 } }&  \\ & {\hspace{2.5cm}\times } & \end{array}

2x^3 - x^2 - 5x - 2 = ( x-2)( 2x^2 + 3x + 1)

= ( x-2)( 2x^2 + 2x + x + 1)

= ( x-2)[ 2x(x+1) + (x+1) ]

= (x-2)( x+1)(2x+1)