Fundamental Concepts and Operations on Algebraic Expressions

Q1: Divide:

i. \(45x^7\) by \(-9x^4\)

Step 1: Write as a fraction: \[ \frac{45x^7}{-9x^4} \] Step 2: Divide the constants: \( \frac{45}{-9} = -5 \)
Step 3: Apply law of exponents: \[ x^{7-4} = x^3 \] Answer: -5x³

ii. \(-60x^3y^2\) by \(-15xy\)

Step 1: Write as a fraction: \[ \frac{-60x^3y^2}{-15xy} \] Step 2: Divide constants: \( \frac{-60}{-15} = 4 \)
Step 3: Apply exponent rules: \[ x^{3-1} = x^2,\quad y^{2-1} = y \] Answer: 4x²y

iii. \(-\frac{3}{4}x^2yz^3\) by \(-\frac{2}{3}x^2yz\)

Step 1: Write as a fraction: \[ \frac{-\frac{3}{4}x^2yz^3}{-\frac{2}{3}x^2yz} \] Step 2: Cancel negatives and divide fractions: \[ \frac{3}{4} \div \frac{2}{3} = \frac{3}{4} \times \frac{3}{2} = \frac{9}{8} \] Step 3: Cancel variables: \[ x^{2-2} = 1,\quad y^{1-1} = 1,\quad z^{3-1} = z^2 \] Answer: \(\frac{9}{8}z^2\)

iv. \(63a^4b^3c^6\) by \(-14a^2b^5c^4\)

Step 1: Write as a fraction: \[ \frac{63a^4b^3c^6}{-14a^2b^5c^4} \] Step 2: Divide constants: \( \frac{63}{-14} = -\frac{9}{2} \)
Step 3: Apply exponent rules: \[ a^{4-2} = a^2,\quad b^{3-5} = b^{-2},\quad c^{6-4} = c^2 \] Answer: \(-\frac{9}{2}a^2b^{-2}c^2\) or \(-\frac{9c^2a^2}{2b^2}\)

Q2: Divide:

i. \(6ax-3cx+15x\) by \(-3x\)

Step 1: Write each term divided by \(-3x\): \[ \frac{6ax}{-3x},\quad \frac{-3cx}{-3x},\quad \frac{15x}{-3x} \] Step 2: Simplify each: \[ \frac{6ax}{-3x} = -2a, \quad \frac{-3cx}{-3x} = c, \quad \frac{15x}{-3x} = -5 \] Answer: \(-2a + c – 5\)

ii. \(10mn^2-15m^2n^2+5m^3n\) by \(-5mn\)

Step 1: Divide each term by \(-5mn\): \[ \frac{10mn^2}{-5mn}, \quad \frac{-15m^2n^2}{-5mn}, \quad \frac{5m^3n}{-5mn} \] Step 2: Simplify constants: \[ \frac{10}{-5} = -2, \quad \frac{-15}{-5} = 3, \quad \frac{5}{-5} = -1 \] Simplify variables using exponent rules: \[ \frac{mn^2}{mn} = n, \quad \frac{m^2 n^2}{mn} = mn, \quad \frac{m^3 n}{mn} = m^2 \] Answer: \(-2n + 3mn – m^2\)

iii. \(14x^3y^4-7x^4y^3-28x^3y^6\) by \(-7x^3y^2\)

Step 1: Divide each term by \(-7x^3 y^2\): \[ \frac{14 x^3 y^4}{-7 x^3 y^2}, \quad \frac{-7 x^4 y^3}{-7 x^3 y^2}, \quad \frac{-28 x^3 y^6}{-7 x^3 y^2} \] Step 2: Simplify constants: \[ \frac{14}{-7} = -2, \quad \frac{-7}{-7} = 1, \quad \frac{-28}{-7} = 4 \] Simplify variables: \[ \frac{x^3}{x^3} = 1, \quad \frac{y^4}{y^2} = y^2, \quad \frac{x^4}{x^3} = x, \quad \frac{y^3}{y^2} = y, \quad \frac{y^6}{y^2} = y^4 \] Answer: \(-2 y^{2} + x y + 4 y^{4}\)

iv. \(18a^6b^3-30a^4b^5+6a^4b^4\) by \(6a^2b^2\)

Step 1: Divide each term by \(6 a^{2} b^{2}\): \[ \frac{18 a^{6} b^{3}}{6 a^{2} b^{2}}, \quad \frac{-30 a^{4} b^{5}}{6 a^{2} b^{2}}, \quad \frac{6 a^{4} b^{4}}{6 a^{2} b^{2}} \] Step 2: Simplify constants: \[ \frac{18}{6} = 3, \quad \frac{-30}{6} = -5, \quad \frac{6}{6} = 1 \] Simplify variables: \[ a^{6-2} = a^{4}, \quad b^{3-2} = b^{1}, \quad a^{4-2} = a^{2}, \quad b^{5-2} = b^{3}, \quad b^{4-2} = b^{2} \] Answer: \(3 a^{4} b – 5 a^{2} b^{3} + a^{2} b^{2}\)

v. \(8a^3b-16a^2b^2-6a^4b^3\) by \(-2a^2b\)

Step 1: Divide each term by \(-2 a^{2} b\): \[ \frac{8 a^{3} b}{-2 a^{2} b}, \quad \frac{-16 a^{2} b^{2}}{-2 a^{2} b}, \quad \frac{-6 a^{4} b^{3}}{-2 a^{2} b} \] Step 2: Simplify constants: \[ \frac{8}{-2} = -4, \quad \frac{-16}{-2} = 8, \quad \frac{-6}{-2} = 3 \] Simplify variables: \[ a^{3-2} = a, \quad b^{1-1} = 1, \quad b^{2-1} = b, \quad a^{4-2} = a^{2}, \quad b^{3-1} = b^{2} \] Answer: \(-4 a + 8 b + 3 a^{2} b^{2}\)

vi. \(\frac{1}{2}p^2q^3-\frac{5}{8}p^3q^2+\frac{1}{4}p^3q^3\) by \(-\frac{1}{4}p^2q^2\)

Step 1: Divide each term by \(-\frac{1}{4} p^{2} q^{2}\): \[ \frac{\frac{1}{2} p^{2} q^{3}}{-\frac{1}{4} p^{2} q^{2}}, \quad \frac{-\frac{5}{8} p^{3} q^{2}}{-\frac{1}{4} p^{2} q^{2}}, \quad \frac{\frac{1}{4} p^{3} q^{3}}{-\frac{1}{4} p^{2} q^{2}} \] Step 2: Simplify constants: \[ \frac{\frac{1}{2}}{-\frac{1}{4}} = -2, \quad \frac{-\frac{5}{8}}{-\frac{1}{4}} = \frac{5}{2}, \quad \frac{\frac{1}{4}}{-\frac{1}{4}} = -1 \] Simplify variables: \[ p^{2-2} = 1, \quad q^{3-2} = q, \quad p^{3-2} = p, \quad q^{2-2} = 1, \quad p^{3-2} = p, \quad q^{3-2} = q \] Answer: \(-2 q + \frac{5}{2} p – p q\)

Q3: Divide:

i. \(\left(x^2+8x+15\right)\) by \(\left(x+5\right)\)

Step 1: Divide \(x^2 + 8x + 15\) by \(x + 5\) using polynomial division:
Divide \(x^2\) by \(x\), quotient term is \(x\).
Multiply \(x \times (x + 5) = x^2 + 5x\).
Subtract: \((x^2 + 8x + 15) – (x^2 + 5x) = 3x + 15\).
Divide \(3x\) by \(x\), quotient term is \(+3\).
Multiply \(3 \times (x + 5) = 3x + 15\).
Subtract: \((3x + 15) – (3x + 15) = 0\).
Answer: \(x + 3\)

ii. \(\left(4x^2+11x-3\right)\) by \(\left(x+3\right)\)

Step 1: Divide \(4x^2 + 11x – 3\) by \(x + 3\):
Divide \(4x^2\) by \(x\), quotient term is \(4x\).
Multiply \(4x \times (x + 3) = 4x^2 + 12x\).
Subtract: \((4x^2 + 11x – 3) – (4x^2 + 12x) = -x – 3\).
Divide \(-x\) by \(x\), quotient term is \(-1\).
Multiply \(-1 \times (x + 3) = -x – 3\).
Subtract: \((-x – 3) – (-x – 3) = 0\).
Answer: \(4x – 1\)

iii. \(\left(3+x-2x^2\right)\) by \(\left(x+1\right)\)

Step 1: Rewrite dividend as \(-2x^2 + x + 3\). Divide by \(x + 1\):
Divide \(-2x^2\) by \(x\), quotient term is \(-2x\).
Multiply \(-2x \times (x + 1) = -2x^2 – 2x\).
Subtract: \((-2x^2 + x + 3) – (-2x^2 – 2x) = 3x + 3\).
Divide \(3x\) by \(x\), quotient term is \(+3\).
Multiply \(3 \times (x + 1) = 3x + 3\).
Subtract: \((3x + 3) – (3x + 3) = 0\).
Answer: \(-2x + 3\)

iv. \(\left(x^3-9x^2+26x-24\right)\) by \(\left(x-4\right)\)

Step 1: Divide \(x^3 – 9x^2 + 26x – 24\) by \(x – 4\):
Divide \(x^3\) by \(x\), quotient term is \(x^2\).
Multiply \(x^2 \times (x – 4) = x^3 – 4x^2\).
Subtract: \((x^3 – 9x^2 + 26x – 24) – (x^3 – 4x^2) = -5x^2 + 26x – 24\).
Divide \(-5x^2\) by \(x\), quotient term is \(-5x\).
Multiply \(-5x \times (x – 4) = -5x^2 + 20x\).
Subtract: \((-5x^2 + 26x – 24) – (-5x^2 + 20x) = 6x – 24\).
Divide \(6x\) by \(x\), quotient term is \(+6\).
Multiply \(6 \times (x – 4) = 6x – 24\).
Subtract: \((6x – 24) – (6x – 24) = 0\).
Answer: \(x^{2} – 5x + 6\)

v. \(\left(x^3-3x^2y+3xy^2-y^3\right)\) by \(\left(x-y\right)\)

Step 1: Divide \(x^{3} – 3x^{2}y + 3xy^{2} – y^{3}\) by \(x – y\):
This is a cubic expansion of \((x – y)^3\), so: \[ \frac{(x – y)^3}{(x – y)} = (x – y)^2 = x^2 – 2xy + y^2 \] Answer: \(x^{2} – 2xy + y^{2}\)

Q4: Divide:

i. \(\left(12x^2+11x+2\right)\) by \(\left(4x+1\right)\)

Step 1: Divide \(12x^2 + 11x + 2\) by \(4x + 1\) using polynomial division:
Divide \(12x^2\) by \(4x\), quotient term is \(3x\).
Multiply \(3x \times (4x + 1) = 12x^2 + 3x\).
Subtract: \((12x^2 + 11x + 2) – (12x^2 + 3x) = 8x + 2\).
Divide \(8x\) by \(4x\), quotient term is \(+2\).
Multiply \(2 \times (4x + 1) = 8x + 2\).
Subtract: \((8x + 2) – (8x + 2) = 0\).
Answer: \(3x + 2\)

ii. \(\left(6x^2+x+15\right)\) by \(\left(2x-3\right)\)

Step 1: Divide \(6x^2 + x – 15\) by \(2x – 3\):
Divide \(6x^2\) by \(2x\), quotient term is \(3x\).
Multiply \(3x \times (2x – 3) = 6x^2 – 9x\).
Subtract: \((6x^2 + x – 15) – (6x^2 – 9x) = 10x – 15\).
Divide \(10x\) by \(2x\), quotient term is \(+5\).
Multiply \(5 \times (2x – 3) = 10x – 15\).
Subtract: \((10x – 15) – (10x – 15) = 0\).
Answer: \(3x + 5\)

iii. \(\left(6x^3+x^2-26x-21\right)\) by \(\left(3x-7\right)\)

Step 1: Divide \(6x^3 + x^2 – 26x – 21\) by \(3x – 7\):
Divide \(6x^3\) by \(3x\), quotient term is \(2x^2\).
Multiply \(2x^2 \times (3x – 7) = 6x^3 – 14x^2\).
Subtract: \((6x^3 + x^2 – 26x – 21) – (6x^3 – 14x^2) = 15x^2 – 26x – 21\).
Divide \(15x^2\) by \(3x\), quotient term is \(5x\).
Multiply \(5x \times (3x – 7) = 15x^2 – 35x\).
Subtract: \((15x^2 – 26x – 21) – (15x^2 – 35x) = 9x – 21\).
Divide \(9x\) by \(3x\), quotient term is \(+3\).
Multiply \(3 \times (3x – 7) = 9x – 21\).
Subtract: \((9x – 21) – (9x – 21) = 0\).
Answer: \(2x^2 + 5x + 3\)

iv. \(\left(12x^2+7xy-12y^2\right)\) by \(\left(3x+4y\right)\)

Step 1: Divide \(12x^2 + 7xy – 12y^2\) by \(3x + 4y\):
Divide \(12x^2\) by \(3x\), quotient term is \(4x\).
Multiply \(4x \times (3x + 4y) = 12x^2 + 16xy\).
Subtract: \((12x^2 + 7xy – 12y^2) – (12x^2 + 16xy) = -9xy – 12y^2\).
Divide \(-9xy\) by \(3x\), quotient term is \(-3y\).
Multiply \(-3y \times (3x + 4y) = -9xy – 12y^2\).
Subtract: \((-9xy – 12y^2) – (-9xy – 12y^2) = 0\).
Answer: \(4x – 3y\)

Q5: Divide:

i. \((x^3 – 2x – 1)\) by \((x^2 – x – 1)\)

Step 1: Divide the first term of dividend by the first term of divisor:
\(\frac{x^3}{x^2} = x\)
Step 2: Multiply \(x\) with the divisor:
\(x(x^2 – x – 1) = x^3 – x^2 – x\)
Step 3: Subtract:
\((x^3 – 2x – 1) – (x^3 – x^2 – x) = x^2 – x – 1\)
Step 4: Divide the first term again:
\(\frac{x^2}{x^2} = 1\)
Step 5: Multiply and subtract:
\(1(x^2 – x – 1) = x^2 – x – 1\)
\((x^2 – x – 1) – (x^2 – x – 1) = 0\)
Answer: Quotient = \(x + 1\), Remainder = 0

ii. \((x^3 – 6x^2 + 11x – 6)\) by \((x^2 – 5x + 6)\)

Step 1: \(\frac{x^3}{x^2} = x\)
Step 2: \(x(x^2 – 5x + 6) = x^3 – 5x^2 + 6x\)
Step 3: Subtract: \(x^3 – 6x^2 + 11x – 6 – (x^3 – 5x^2 + 6x)\)
Result: \(-x^2 + 5x – 6\)
Step 4: \(\frac{-x^2}{x^2} = -1\)
Step 5: \(-1(x^2 – 5x + 6) = -x^2 + 5x – 6\)
Subtracting gives: 0
Answer: Quotient = \(x – 1\), Remainder = 0

iii. \((6x^5 + 4x^4 – 3x^3 – 1)\) by \((3x^2 – x + 1)\)

Step 1: \(\frac{6x^5}{3x^2} = 2x^3\)
Step 2: Multiply:
\(2x^3(3x^2 – x + 1) = 6x^5 – 2x^4 + 2x^3\)
Step 3: Subtract:
\((6x^5 + 4x^4 – 3x^3) – (6x^5 – 2x^4 + 2x^3) = 6x^4 – 5x^3\)
Bring down: \(-1\), New dividend: \(6x^4 – 5x^3 – 1\)
Step 4: \(\frac{6x^4}{3x^2} = 2x^2\)
Multiply: \(2x^2(3x^2 – x + 1) = 6x^4 – 2x^3 + 2x^2\)
Subtract: \((6x^4 – 5x^3) – (6x^4 – 2x^3) = -3x^3 – 2x^2\)
Bring down: \(-1\), New dividend: \(-3x^3 – 2x^2 – 1\)
Step 5: \(\frac{-3x^3}{3x^2} = -x\)
Multiply: \(-x(3x^2 – x + 1) = -3x^3 + x^2 – x\)
Subtract: \(-3x^3 – 2x^2 – 1 – (-3x^3 + x^2 – x) = -3x^2 + x – 1\)
Step 6: \(\frac{-3x^2}{3x^2} = -1\)
Multiply: \(-1(3x^2 – x + 1) = -3x^2 + x – 1\)
Subtract: gives 0
Answer: Quotient = \(2x^3 + 2x^2 – x – 1\), Remainder = 0

iv. \((6x^5 – 28x^3 + 3x^2 + 30x – 9)\) by \((2x^2 – 6)\)

Step 1: \(\frac{6x^5}{2x^2} = 3x^3\)
Multiply: \(3x^3(2x^2 – 6) = 6x^5 – 18x^3\)
Subtract: \((6x^5 – 28x^3) – (6x^5 – 18x^3) = -10x^3\)
Bring down: \(+3x^2\), New dividend: \(-10x^3 + 3x^2\)
Step 2: \(\frac{-10x^3}{2x^2} = -5x\)
Multiply: \(-5x(2x^2 – 6) = -10x^3 + 30x\)
Subtract: \(-10x^3 + 3x^2 + 30x – (-10x^3 + 30x) = 3x^2\)
Bring down: \(-9\), New dividend: \(3x^2 – 9\)
Step 3: \(\frac{3x^2}{2x^2} = \frac{3}{2}\)
Multiply: \(\frac{3}{2}(2x^2 – 6) = 3x^2 – 9\)
Subtract: \(3x^2 – 9 – (3x^2 – 9) = 0\)
Answer: Quotient = \(3x^3 – 5x + \frac{3}{2}\), Remainder = 0

Q6: Divide:

i. \((x^3 + 27)\) by \((x + 3)\)

Step 1: Recognize the identity for sum of cubes:
\(a^3 + b^3 = (a + b)(a^2 – ab + b^2)\)
Step 2: Here, \(x^3 + 27 = x^3 + 3^3\)
Step 3: Apply the identity:
\(x^3 + 3^3 = (x + 3)(x^2 – 3x + 9)\)
Step 4: Now divide:
\(\frac{x^3 + 27}{x + 3} = x^2 – 3x + 9\)
Answer: \(x^2 – 3x + 9\)

ii. \((x^4 – 81)\) by \((x + 3)\)

Step 1: First factor \(x^4 – 81\)
This is a difference of squares:
\(x^4 – 81 = (x^2)^2 – (9)^2 = (x^2 – 9)(x^2 + 9)\)
Step 2: Factor further:
\(x^2 – 9 = (x – 3)(x + 3)\)
Step 3: So full factorization:
\(x^4 – 81 = (x – 3)(x + 3)(x^2 + 9)\)
Step 4: Now divide by \((x + 3)\):
\(\frac{(x – 3)(x + 3)(x^2 + 9)}{x + 3} = (x – 3)(x^2 + 9) = x^3 – 3x^2 + 9x – 27\)
Answer: \((x – 3)(x^2 + 9) = x^3 – 3x^2 + 9x – 27\)

iii. \((27x^3 – 8)\) by \((3x – 2)\)

Step 1: Recognize identity for difference of cubes:
\(a^3 – b^3 = (a – b)(a^2 + ab + b^2)\)
Step 2: Express in cube form:
\(27x^3 = (3x)^3\) and \(8 = 2^3\)
So, \(27x^3 – 8 = (3x)^3 – 2^3\)
Step 3: Apply identity:
\((3x – 2)((3x)^2 + (3x)(2) + 2^2) = (3x – 2)(9x^2 + 6x + 4)\)
Step 4: Divide:
\(\frac{27x^3 – 8}{3x – 2} = 9x^2 + 6x + 4\)
Answer: \(9x^2 + 6x + 4\)

iv. \((x^6 – 8)\) by \((x^2 – 2)\)

Step 1: Let’s factor numerator using identity:
\(x^6 – 8 = (x^2)^3 – 2^3 =\) difference of cubes
Step 2: Apply identity:
\((x^2 – 2)(x^4 + 2x^2 + 4)\)
Step 3: Divide:
\(\frac{(x^2 – 2)(x^4 + 2x^2 + 4)}{x^2 – 2} = x^4 + 2x^2 + 4\)
Answer: \(x^4 + 2x^2 + 4\)

v. \((x^6 – y^6)\) by \((x – y)\)

Step 1: Use identity:
\(a^6 – b^6 = (a – b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)\)
Step 2: Apply for \(x^6 – y^6\):
\((x – y)(x^5 + x^4y + x^3y^2 + x^2y^3 + xy^4 + y^5)\)
Step 3: Divide:
\(\frac{(x – y)(x^5 + x^4y + x^3y^2 + x^2y^3 + xy^4 + y^5)}{x – y}\)
Answer: \(x^5 + x^4y + x^3y^2 + x^2y^3 + xy^4 + y^5\)

vi. \((16x^4 – 81y^4)\) by \((2x – 3y)\)

Step 1: Recognize this as a difference of squares:
\(16x^4 = (4x^2)^2\) and \(81y^4 = (9y^2)^2\)
Step 2: So:
\(16x^4 – 81y^4 = (4x^2 – 9y^2)(4x^2 + 9y^2)\)
Step 3: Now factor \(4x^2 – 9y^2 = (2x – 3y)(2x + 3y)\)
Step 4: So complete factorization:
\((2x – 3y)(2x + 3y)(4x^2 + 9y^2)\)
Step 5: Divide:
\(\frac{(2x – 3y)(2x + 3y)(4x^2 + 9y^2)}{2x – 3y} = (2x + 3y)(4x^2 + 9y^2) = 8x^3 + 12x^2y + 18xy^2 + 27y^3\)
Answer: \((2x + 3y)(4x^2 + 9y^2) = 8x^3 + 12x^2y + 18xy^2 + 27y^3\)

Q7: Find the quotient and remainder when:

i. \(\left(6a^2-31a+47\right)\) is divided by \(\left(2a-5\right)\)

Step 1: Divide \(\left(6a^2 – 31a + 47\right)\) by \(\left(2a – 5\right)\) using long division:
Divide the leading term \(6a^2\) by \(2a\): \(3a\).
Multiply divisor by \(3a\): \(3a \times (2a – 5) = 6a^2 – 15a\).
Subtract: \((6a^2 – 31a + 47) – (6a^2 – 15a) = -16a + 47\).
Divide \(-16a\) by \(2a\): \(-8\).
Multiply divisor by \(-8\): \(-8 \times (2a – 5) = -16a + 40\).
Subtract: \((-16a + 47) – (-16a + 40) = 7\).
Answer: Quotient = \(3a – 8\), Remainder = 7

ii. \(\left(2x^4-x^3+10x^2+8x-5\right)\) is divided by \(\left(x^2-x+6\right)\)

Step 1: Divide \(\left(2x^4 – x^3 + 10x^2 + 8x – 5\right)\) by \(\left(x^2 – x + 6\right)\) using polynomial division:
Divide \(2x^4\) by \(x^2\): \(2x^2\).
Multiply divisor by \(2x^2\): \(2x^4 – 2x^3 + 12x^2\).
Subtract: \((2x^4 – x^3 + 10x^2) – (2x^4 – 2x^3 + 12x^2) = x^3 – 2x^2\).
Bring down remaining terms: \(x^3 – 2x^2 + 8x – 5\).
Divide \(x^3\) by \(x^2\): \(x\).
Multiply divisor by \(x\): \(x^3 – x^2 + 6x\).
Subtract: \((x^3 – 2x^2 + 8x) – (x^3 – x^2 + 6x) = -x^2 + 2x\).
Bring down \(-5\).
Divide \(-x^2\) by \(x^2\): \(-1\).
Multiply divisor by \(-1\): \(-x^2 + x – 6\).
Subtract: \((-x^2 + 2x – 5) – (-x^2 + x – 6) = x + 1\).
Answer: Quotient = \(2x^2 + x – 1\), Remainder = \(x + 1\)

iii. \(\left(3t^5+7t^4-11t^3+8t^2-32t+5\right)\) is divided by (\left(2+3t+t^2\right)\)

Step 1: Divide \(\left(3t^5 + 7t^4 – 11t^3 + 8t^2 – 32t + 5\right)\) by \(\left(2 + 3t + t^2\right)\) using polynomial long division:
Arrange divisor in descending order: \(t^2 + 3t + 2\).
Divide \(3t^5\) by \(t^2\): \(3t^3\).
Multiply divisor by \(3t^3\): \(3t^5 + 9t^4 + 6t^3\).
Subtract: \((3t^5 + 7t^4 – 11t^3) – (3t^5 + 9t^4 + 6t^3) = -2t^4 -17t^3\).
Bring down \(+8t^2\).
Divide \(-2t^4\) by \(t^2\): \(-2t^2\).
Multiply divisor by \(-2t^2\): \(-2t^4 – 6t^3 – 4t^2\).
Subtract: \((-2t^4 – 17t^3 + 8t^2) – (-2t^4 – 6t^3 – 4t^2) = -11t^3 + 12t^2\).
Bring down \(-32t\).
Divide \(-11t^3\) by \(t^2\): \(-11t\).
Multiply divisor by \(-11t\): \(-11t^3 – 33t^2 – 22t\).
Subtract: \((-11t^3 + 12t^2 – 32t) – (-11t^3 – 33t^2 – 22t) = 45t^2 – 10t\).
Bring down \(+5\).
Divide \(45t^2\) by \(t^2\): \(45\).
Multiply divisor by \(45\): \(45t^2 + 135t + 90\).
Subtract: \((45t^2 – 10t + 5) – (45t^2 + 135t + 90) = -145t – 85\).
Since degree of remainder < degree divisor, division ends.
Answer: Quotient = \(3t^3 – 2t^2 – 11t + 45\), Remainder = \(-145t – 85\)

iv. \(\left(x^6+3x^2+10\right)\) is divided by \(\left(x^3+1\right)\)

Step 1: Divide \(\left(x^6 + 3x^2 + 10\right)\) by \(\left(x^3 + 1\right)\) using polynomial long division:
Divide \(x^6\) by \(x^3\): \(x^3\).
Multiply divisor by \(x^3\): \(x^6 + x^3\).
Subtract: \((x^6 + 3x^2 + 10) – (x^6 + x^3) = -x^3 + 3x^2 + 10\).
Divide \(-x^3\) by \(x^3\): \(-1\).
Multiply divisor by \(-1\): \(-x^3 – 1\).
Subtract: \((-x^3 + 3x^2 + 10) – (-x^3 – 1) = 3x^2 + 11\).
Degree of remainder \(3x^2 + 11\) is less than divisor degree, so division ends.
Answer: Quotient = \(x^3 – 1\), Remainder = \(3x^2 + 11\)

Q8: Show by division method that \(\left(2a^2-a+3\right)\) is a factor of \((6a^5-a^4+4a^3-5a^2-a-15\).

Step 1: We divide the polynomial \(6a^5 – a^4 + 4a^3 – 5a^2 – a – 15\) by \(2a^2 – a + 3\) using polynomial long division.
Step 2: Divide the leading term \(6a^5\) by \(2a^2\) to get the first term of quotient: \(3a^3\).
Multiply divisor by \(3a^3\):
\(3a^3 \times (2a^2 – a + 3) = 6a^5 – 3a^4 + 9a^3\).
Subtract:
\(\left(6a^5 – a^4 + 4a^3\right) – \left(6a^5 – 3a^4 + 9a^3\right) = 2a^4 – 5a^3\).
Bring down the remaining terms: \(2a^4 – 5a^3 – 5a^2 – a – 15\).
Step 3: Divide \(2a^4\) by \(2a^2\) to get \(a^2\).
Multiply divisor by \(a^2\):
\(a^2 \times (2a^2 – a + 3) = 2a^4 – a^3 + 3a^2\).
Subtract:
\(\left(2a^4 – 5a^3 – 5a^2\right) – \left(2a^4 – a^3 + 3a^2\right) = -4a^3 – 8a^2\).
Bring down remaining terms: \(-4a^3 – 8a^2 – a – 15\).
Step 4: Divide \(-4a^3\) by \(2a^2\) to get \(-2a\).
Multiply divisor by \(-2a\):
\(-2a \times (2a^2 – a + 3) = -4a^3 + 2a^2 – 6a\).
Subtract:
\(\left(-4a^3 – 8a^2 – a\right) – \left(-4a^3 + 2a^2 – 6a\right) = -10a^2 + 5a\).
Bring down \(-15\): \(-10a^2 + 5a – 15\).
Step 5: Divide \(-10a^2\) by \(2a^2\) to get \(-5\).
Multiply divisor by \(-5\):
\(-5 \times (2a^2 – a + 3) = -10a^2 + 5a – 15\).
Subtract:
\(\left(-10a^2 + 5a – 15\right) – \left(-10a^2 + 5a – 15\right) = 0\).
Step 6: Since the remainder is zero, the divisor \(2a^2 – a + 3\) is a factor of the given polynomial.
Answer: Quotient = \(3a^3 + a^2 – 2a – 5\), Remainder = 0.
Thus, \(2a^2 – a + 3\) is a factor of the polynomial \(6a^5 – a^4 + 4a^3 – 5a^2 – a – 15\).

Q9: What must be subtracted from \(8x^4+14x^3-2x^2+7x-8\) so that the resulting polynomial is exactly divisible by \(\left(4x^2+3x-2\right)\)?

Step 1: Let the polynomial to be subtracted be \(P(x)\).
We want:
\(8x^4 + 14x^3 – 2x^2 + 7x – 8 – P(x) \text{ is divisible by } (4x^2 + 3x – 2)\).
Step 2: Divide \(8x^4 + 14x^3 – 2x^2 + 7x – 8\) by \(4x^2 + 3x – 2\) using polynomial long division:
Step 3: Divide leading term \(8x^4\) by \(4x^2\) to get \(2x^2\).
Multiply divisor by \(2x^2\):
\(2x^2 \times (4x^2 + 3x – 2) = 8x^4 + 6x^3 – 4x^2\).
Subtract:
\((8x^4 + 14x^3 – 2x^2) – (8x^4 + 6x^3 – 4x^2) = 8x^3 + 2x^2\).
Bring down remaining terms: \(8x^3 + 2x^2 + 7x – 8\).
Step 4: Divide \(8x^3\) by \(4x^2\) to get \(2x\).
Multiply divisor by \(2x\):
\(2x \times (4x^2 + 3x – 2) = 8x^3 + 6x^2 – 4x\).
Subtract:
\((8x^3 + 2x^2 + 7x) – (8x^3 + 6x^2 – 4x) = -4x^2 + 11x\).
Bring down \(-8\): \(-4x^2 + 11x – 8\).
Step 5: Divide \(-4x^2\) by \(4x^2\) to get \(-1\).
Multiply divisor by \(-1\):
\(-1 \times (4x^2 + 3x – 2) = -4x^2 – 3x + 2\).
Subtract:
\((-4x^2 + 11x – 8) – (-4x^2 – 3x + 2) = 14x – 10\).
Step 6: The remainder after division is \(14x – 10\).
For the polynomial to be exactly divisible, remainder must be zero.
Hence, \[ 8x^4 + 14x^3 – 2x^2 + 7x – 8 – P(x) = (4x^2 + 3x – 2) \times Q(x), \] where \(Q(x) = 2x^2 + 2x – 1\).
So, \[ P(x) = \text{remainder} = 14x – 10. \]Answer: The polynomial that must be subtracted is \(14x – 10\).

Q10: The length and breadth of a rectangle are \(\left(a+5b\right)\) units and \(\left(7a-b\right)\) units respectively. The perimeter of this rectangle is equal to the perimeter of a square. Find how much is the area of the less than that the square?

Step 1: Write the perimeter of the rectangle. \[ P_{rectangle} = 2 \times \left(\text{length} + \text{breadth}\right) = 2 \times \left((a + 5b) + (7a – b)\right) \\ = 2 \times (8a + 4b) = 16a + 8b \]Step 2: Let the side of the square be \(s\). Then, perimeter of the square: \[ P_{square} = 4s \] Given that \[ P_{rectangle} = P_{square} \\ \Rightarrow 16a + 8b = 4s \\ \Rightarrow s = \frac{16a + 8b}{4} = 4a + 2b \]Step 3: Find the area of the rectangle: \[ A_{rectangle} = \text{length} \times \text{breadth} = (a + 5b)(7a – b) \] Multiply: \[ = a \times 7a + a \times (-b) + 5b \times 7a + 5b \times (-b) = 7a^2 – a b + 35ab – 5b^2 \] Simplify: \[ = 7a^2 + 34ab – 5b^2 \]Step 4: Find the area of the square: \[ A_{square} = s^2 = (4a + 2b)^2 = (4a)^2 + 2 \times 4a \times 2b + (2b)^2 \\ = 16a^2 + 16ab + 4b^2 \]Step 5: Find how much the area of rectangle is less than that of square: \[ A_{square} – A_{rectangle} = (16a^2 + 16ab + 4b^2) – (7a^2 + 34ab – 5b^2) \\ = (16a^2 – 7a^2) + (16ab – 34ab) + (4b^2 + 5b^2) = 9a^2 – 18ab + 9b^2 \]Answer: The area of the rectangle is less than the area of the square by \(9a^2 – 18ab + 9b^2\) square units.

Q11: If a sum of ₹\(\left(16x^3-46x^2+39x-9\right)\) is to be divided equally among \(\left(8x-3\right)\) persons, find the amount recieved by each person.

Step 1: The amount received by each person = \[ \frac{\text{Total sum}}{\text{Number of persons}} = \frac{16x^3 – 46x^2 + 39x – 9}{8x – 3} \] We need to perform polynomial division.
Step 2: Divide \(16x^3\) by \(8x\) to get \(2x^2\).
Multiply divisor by \(2x^2\): \[ 2x^2 \times (8x – 3) = 16x^3 – 6x^2 \] Subtract: \[ (16x^3 – 46x^2) – (16x^3 – 6x^2) = -40x^2 \] Bring down remaining terms: \[ -40x^2 + 39x – 9 \]Step 3: Divide \(-40x^2\) by \(8x\) to get \(-5x\).
Multiply divisor by \(-5x\): \[ -5x \times (8x – 3) = -40x^2 + 15x \] Subtract: \[ (-40x^2 + 39x) – (-40x^2 + 15x) = 24x \] Bring down \(-9\): \[ 24x – 9 \]Step 4: Divide \(24x\) by \(8x\) to get \(3\).
Multiply divisor by \(3\): \[ 3 \times (8x – 3) = 24x – 9 \] Subtract: \[ (24x – 9) – (24x – 9) = 0 \]Step 5: The quotient is: \[ 2x^2 – 5x + 3 \] and remainder is 0.
So, \[ \frac{16x^3 – 46x^2 + 39x – 9}{8x – 3} = 2x^2 – 5x + 3 \]Answer: Each person receives ₹\((2x^2 – 5x + 3)\).

Q12: The product of two numbers is \(\left(x^6-y^6\right)\). If one of the numbers is \((x-y)\), then find the other.

Step 1: Given that the product of two numbers is: \[ x^6 – y^6 \] and one of the numbers is: \[ x – y \] We need to find the other number, say \(N\).
Step 2: Write the relation: \[ (x – y) \times N = x^6 – y^6 \\ \Rightarrow N = \frac{x^6 – y^6}{x – y} \]Step 3: Use the formula for difference of sixth powers: \[ x^6 – y^6 = (x – y)(x^5 + x^4 y + x^3 y^2 + x^2 y^3 + x y^4 + y^5) \] Therefore, \[ N = \frac{(x – y)(x^5 + x^4 y + x^3 y^2 + x^2 y^3 + x y^4 + y^5)}{x – y} = x^5 + x^4 y + x^3 y^2 + x^2 y^3 + x y^4 + y^5 \]Answer: The other number is \(x^5 + x^4 y + x^3 y^2 + x^2 y^3 + x y^4 + y^5\).

Q13: Divide \(6x^4-13x^3+2x^2+22x-24\) by the product of \(2x-3\) and \(x^2-2x+2\).

Step 1: Write the divisor as product: \[ D = (2x – 3)(x^2 – 2x + 2) \]Step 2: First, expand the divisor: \[ (2x – 3)(x^2 – 2x + 2) = 2x \times x^2 + 2x \times (-2x) + 2x \times 2 – 3 \times x^2 – 3 \times (-2x) – 3 \times 2 \\ = 2x^3 – 4x^2 + 4x – 3x^2 + 6x – 6 \\ = 2x^3 – 7x^2 + 10x – 6 \]Step 3: Now, divide the polynomial \(6x^4 – 13x^3 + 2x^2 + 22x – 24\) by the divisor polynomial \(2x^3 – 7x^2 + 10x – 6\).
Let the quotient be \(Q(x)\) and remainder \(R(x)\).
Step 4: Divide leading terms: \[ \frac{6x^4}{2x^3} = 3x \] Multiply divisor by \(3x\): \[ 3x \times (2x^3 – 7x^2 + 10x – 6) = 6x^4 – 21x^3 + 30x^2 – 18x \] Subtract: \[ (6x^4 – 13x^3 + 2x^2 + 22x – 24) – (6x^4 – 21x^3 + 30x^2 – 18x) = (0) + (8x^3) + (-28x^2) + (40x) – 24 \]Step 5: Divide leading terms: \[ \frac{8x^3}{2x^3} = 4 \] Multiply divisor by \(4\): \[ 4 \times (2x^3 – 7x^2 + 10x – 6) = 8x^3 – 28x^2 + 40x – 24 \] Subtract: \[ (8x^3 – 28x^2 + 40x – 24) – (8x^3 – 28x^2 + 40x – 24) = 0 \]Step 6: The quotient is: \[ Q(x) = 3x + 4 \] and the remainder is zero.
Answer: The quotient is \(3x + 4\) and the remainder is 0.