Algebraic Expressions

Q1: Multiple Choice Type

i. \((9x^4 – 8x^3 – 12) \times (3x)\) is equal to:

Step 1: Distribute \(3x\) to each term inside the bracket.
\((9x^4 \times 3x) + (-8x^3 \times 3x) + (-12 \times 3x)\)
Step 2: Multiply each term.
\(27x^5 – 24x^4 – 36x\)
Answer: b. \(27x^5 – 24x^4 – 36x\)

ii. \(\left(9x^4 – 12x^3 – 18x\right) \div (3x)\) is equal to:

Step 1: Divide each term by \(3x\)
\(\frac{9x^4}{3x} – \frac{12x^3}{3x} – \frac{18x}{3x}\)
Step 2: Simplify the terms:
\(3x^3 – 4x^2 – 6\)
Answer: c. \(3x^3 – 4x^2 – 6\)

iii. \(\left(\frac{10}{3}xy^2z\right) \times \left(-\frac{9}{5}x^2z\right)\) is equal to:

Step 1: Multiply the coefficients:
\(\frac{10}{3} \times -\frac{9}{5} = -6\)
Step 2: Multiply the variables:
\(x \times x^2 = x^3,\quad y^2,\quad z \times z = z^2\)
Step 3: Final expression:
\(-6x^3y^2z^2\)
Answer: a. \(-6x^3y^2z^2\)

iv. \((x^3 + y^2) \times 10x^2\) is equal to:

Step 1: Distribute \(10x^2\):
\((x^3 \times 10x^2) + (y^2 \times 10x^2)\)
Step 2: Multiply each term:
\(10x^5 + 10x^2y^2\)
Answer: b. \(10x^5 + 10x^2y^2\)

v. \((a^3 – b^3) \div (a – b)\) is equal to:

Step 1: Use the identity:
\(a^3 – b^3 = (a – b)(a^2 + ab + b^2)\)
Step 2: Cancel the common factor \((a – b)\):
\(\frac{(a – b)(a^2 + ab + b^2)}{(a – b)} = a^2 + ab + b^2\)
Answer: a. \(a^2 + ab + b^2\)

Q2: Multiply:

i. \(8ab^2\) by \(-4a^3b^4\)

Step 1: Multiply the coefficients: 8 × (-4) = -32
Step 2: Multiply variables: \(a × a^3 = a^4\), \(b^2 × b^4 = b^6\)
Answer: \(-32a^4b^6\)

ii. \(\frac{2}{3}ab\) by \(-\frac{1}{4}a^2b\)

Step 1: Multiply coefficients: \(\frac{2}{3} × -\frac{1}{4} = -\frac{2}{12} = -\frac{1}{6}\)
Step 2: Multiply variables: \(a × a^2 = a^3\), \(b × b = b^2\)
Answer: \(-\frac{1}{6}a^3b^2\)

iii. \(-5cd^2\) by \(-5cd^2\)

Step 1: Multiply coefficients: \(-5 × -5 = 25\)
Step 2: \(c × c = c^2\), \(d^2 × d^2 = d^4\)
Answer: \(25c^2d^4\)

iv. \(4a\) by \(6a+7\)

Step 1: Distribute: \(4a × 6a = 24a^2\), \(4a × 7 = 28a\)
Answer: \(24a^2 + 28a\)

v. \(-8x\) by \(4 – 2x – x^2\)

Step 1: Distribute:
\(-8x × 4 = -32x\)
\(-8x × -2x = +16x^2\)
\(-8x × -x^2 = +8x^3\)
Answer: \(8x^3 + 16x^2 – 32x\)

vi. \(2a^2 – 5a – 4\) by \(-3a\)

Step 1: Distribute:
\(-3a × 2a^2 = -6a^3\)
\(-3a × -5a = +15a^2\)
\(-3a × -4 = +12a\)
Answer: \(-6a^3 + 15a^2 + 12a\)

vii. \(x+4\) by \(x-5\)

Step 1: Apply distributive property:
\(x × x = x^2\)
\(x × -5 = -5x\)
\(4 × x = 4x\)
\(4 × -5 = -20\)
Step 2: Combine like terms: \(-5x + 4x = -x\)
Answer: \(x^2 – x – 20\)

viii. \(5a-1\) by \(7a-3\)

Step 1: Multiply using distributive law:
\(5a × 7a = 35a^2\)
\(5a × -3 = -15a\)
\(-1 × 7a = -7a\)
\(-1 × -3 = +3\)
Step 2: Combine like terms: \(-15a – 7a = -22a\)
Answer: \(35a^2 – 22a + 3\)

ix. \(12a + 5b\) by \(7a – b\)

Step 1: Use distributive property:
  \((12a + 5b)(7a – b) = 12a \cdot 7a + 12a \cdot (-b) + 5b \cdot 7a + 5b \cdot (-b)\)
Step 2: Multiply each term:
  \(84a^2 – 12ab + 35ab – 5b^2\)
Step 3: Combine like terms:
  \(84a^2 + 23ab – 5b^2\)
Answer: \(84a^2 + 23ab – 5b^2\)

x. \(x^2 + x + 1\) by \(1 – x\)

Step 1: Apply distributive property:
  \((x^2 + x + 1)(1 – x) = x^2 \cdot 1 + x \cdot 1 + 1 \cdot 1 – x^2 \cdot x – x \cdot x – 1 \cdot x\)
Step 2: Multiply:
  \(x^2 + x + 1 – x^3 – x^2 – x\)
Step 3: Combine like terms:
  \(-x^3 + 1\)
Answer: \(-x^3 + 1\)

xi. \(2m^2 – 3m – 1\) and \(4m^2 – m – 1\)

Step 1: Use distributive property: Multiply each term in the first bracket with every term in the second:
  \((2m^2)(4m^2 – m – 1) = 8m^4 – 2m^3 – 2m^2\)
  \((-3m)(4m^2 – m – 1) = -12m^3 + 3m^2 + 3m\)
  \((-1)(4m^2 – m – 1) = -4m^2 + m + 1\)
Step 2: Add all terms:
  \(8m^4 – 2m^3 – 2m^2 – 12m^3 + 3m^2 + 3m – 4m^2 + m + 1\)
Step 3: Combine like terms:
  \(8m^4 – 14m^3 – 3m^2 + 4m + 1\)
Answer: \(8m^4 – 14m^3 – 3m^2 + 4m + 1\)

xii. \(a^2\), \(ab\) and \(b^2\)

Step 1: Multiply all terms:
  \(a^2 \cdot ab \cdot b^2 = a^3b^3\)
Answer: \(a^3b^3\)

xiii. \(abx\), \(-3a^2x\), and \(7b^2x^3\)

Step 1: Multiply all terms:
  \(abx \cdot (-3a^2x) \cdot 7b^2x^3\)
Step 2: Group and simplify:
  \(-3 \cdot 7 \cdot a \cdot a^2 \cdot b \cdot b^2 \cdot x \cdot x \cdot x^3 = -21a^3b^3x^5\)
Answer: \(-21a^3b^3x^5\)

xiv. \(-3bx\), \(-5xy\), and \(-7b^3y^2\)

Step 1: Multiply all constants: \((-3)(-5)(-7) = -105\)
Step 2: Multiply variables:
  \(b \cdot x \cdot x \cdot y \cdot b^3 \cdot y^2 = b^4x^2y^3\)
Answer: \(-105b^4x^2y^3\)

xv. \(-\frac{3}{2}x^5y^3\) and \(\frac{4}{9}a^2x^3y\)

Step 1: Multiply coefficients:
  \(-\frac{3}{2} \cdot \frac{4}{9} = -\frac{12}{18} = -\frac{2}{3}\)
Step 2: Multiply variables:
  \(x^5 \cdot x^3 = x^8,\quad y^3 \cdot y = y^4\)
Answer: \(-\frac{2}{3}a^2x^8y^4\)

xvi. \(-\frac{2}{3}a^7b^2\) and \(-\frac{9}{4}ab^5\)

Step 1: Multiply coefficients:
  \(-\frac{2}{3} \cdot -\frac{9}{4} = \frac{18}{12} = \frac{3}{2}\)
Step 2: Multiply variables:
  \(a^7 \cdot a = a^8,\quad b^2 \cdot b^5 = b^7\)
Answer: \(\frac{3}{2}a^8b^7\)

xvii. \(2a^3 – 3a^2b\) and \(-\frac{1}{2}ab^2\)

Step 1: Distribute:
  \(2a^3 \cdot -\frac{1}{2}ab^2 = -a^4b^2\)
  \(-3a^2b \cdot -\frac{1}{2}ab^2 = \frac{3}{2}a^3b^3\)
Answer: \(-a^4b^2 + \frac{3}{2}a^3b^3\)

xviii. \(2x + \frac{1}{2}y\) and \(2x – \frac{1}{2}y\)

Step 1: Recognize identity: \((a + b)(a – b) = a^2 – b^2\)
  Here, \(a = 2x,\quad b = \frac{1}{2}y\)
Step 2: Apply identity:
  \((2x)^2 – \left(\frac{1}{2}y\right)^2 = 4x^2 – \frac{1}{4}y^2\)
Answer: \(4x^2 – \frac{1}{4}y^2\)

Q3: Multiply

i. \(5x^2 – 8xy + 6y^2 – 3\) by \(-3xy\)

Step 1: Distribute \(-3xy\) to each term inside the bracket
= \(-3xy \times 5x^2\) + \(-3xy \times (-8xy)\) + \(-3xy \times 6y^2\) + \(-3xy \times (-3)\)
Step 2: Multiply each term
= \(-15x^3y + 24x^2 y^2 + (-18x y^3) + 9xy\)
Answer: -15x³y + 24x²y² – 18xy³ + 9xy

ii. \(3 – \frac{2}{3}xy + \frac{5}{7}xy^2 – \frac{16}{21}x^2y\) by \(-21x^2y^2\)

Step 1: Multiply each term
= \(-21x^2y^2 \times 3\) + \(-21x^2y^2 \times (-\frac{2}{3}xy)\) + \(-21x^2y^2 \times \frac{5}{7}xy^2\) + \(-21x^2y^2 \times (-\frac{16}{21}x^2y)\)
Step 2: Solve step-by-step
= \(-63x^2y^2 + 14x^3y^3 – 15x^3y^4 + 16x^4y^3\)
Answer: -63x²y² + 14x³y³ – 15x³y⁴ + 16x⁴y³

iii. \(6x^3 – 5x + 10\) by \(4 – 3x^2\)

Step 1: Distribute both terms
= \(6x^3 \times 4 + 6x^3 \times (-3x^2)\) + \((-5x \times 4 + (-5x \times -3x^2))\) + \(10 \times 4 + 10 \times (-3x^2)\)
Step 2: Multiply
= \(24x^3 – 18x^5 – 20x + 15x^3 + 40 – 30x^2\)
Step 3: Arrange in standard form
= \(-18x^5 + 39x^3 – 30x^2 – 20x + 40\)
Answer: -18x⁵ + 39x³ – 30x² – 20x + 40

iv. \(2y – 4y^3 + 6y^5\) by \(y^2 + y – 3\)

Step 1: Use distributive property:
Each term of first polynomial multiplied by entire second polynomial.
Step 2: Multiply
2y × (y² + y – 3) = 2y³ + 2y² – 6y
-4y³ × (y² + y – 3) = -4y⁵ – 4y⁴ + 12y³
6y⁵ × (y² + y – 3) = 6y⁷ + 6y⁶ – 18y⁵
Step 3: Combine like terms
= 6y⁷ + 6y⁶ – 22y⁵ – 4y⁴ + 14y³ + 2y² – 6y
Answer: 6y⁷ + 6y⁶ – 22y⁵ – 4y⁴ + 14y³ + 2y² – 6y

v. \(5p^2 + 25pq + 4q^2\) by \(2p^2 – 2pq + 3q^2\)

Step 1: Use distributive property (FOIL-like)
First: 5p² × \(2p^2 – 2pq + 3q^2\) = 10p⁴ – 10p³q + 15p²q²
Next: 25pq × \(2p^2 – 2pq + 3q^2\) = 50p³q – 50p²q² + 75pq³
Next: 4q² × \(2p^2 – 2pq + 3q^2\) = 8p²q² – 8pq³ + 12q⁴
Step 2: Combine like terms
= 10p⁴ + 40p³q – 27p²q² + 67pq³ + 12q⁴
Answer: 10p⁴ + 40p³q – 27p²q² + 67pq³ + 12q⁴

Q4: Simplify:

i. \((7x – 8)(3x + 2)\)

Step 1: Apply distributive property (FOIL method):
= \(7x \cdot 3x + 7x \cdot 2 – 8 \cdot 3x – 8 \cdot 2\)
= \(21x^2 + 14x – 24x – 16\)
Step 2: Combine like terms:
= \(21x^2 – 10x – 16\)
Answer: 21x² – 10x – 16

ii. \((px – q)(px + q)\)

Step 1: Use identity \((a – b)(a + b) = a^2 – b^2\)
= \((px)^2 – q^2\)
= \(p^2x^2 – q^2\)
Answer: p²x² – q²

iii. \((5a + 5b – c)(2b – 3c)\)

Step 1: Multiply each term in first bracket with second bracket:
= \(5a \cdot 2b + 5a \cdot (-3c) + 5b \cdot 2b + 5b \cdot (-3c) – c \cdot 2b – c \cdot (-3c)\)
= \(10ab – 15ac + 10b^2 – 15bc – 2bc + 3c^2\)
Step 2: Combine like terms:
= \(10ab – 15ac + 10b^2 – 17bc + 3c^2\)
Answer: 10ab – 15ac + 10b² – 17bc + 3c²

iv. \((4x – 5y)(5x – 4y)\)

Step 1: Multiply using distributive property:
= \(4x \cdot 5x + 4x \cdot (-4y) – 5y \cdot 5x – 5y \cdot (-4y)\)
= \(20x^2 – 16xy – 25xy + 20y^2\)
Step 2: Combine like terms:
= \(20x^2 – 41xy + 20y^2\)
Answer: 20x² – 41xy + 20y²

v. \((3y + 4z)(3y – 4z) + (2y + 7z)(y + z)\)

Step 1: Apply identity on first part:
= \((3y)^2 – (4z)^2\)
= \(9y^2 – 16z^2\)
Step 2: Multiply second part:
= \(2y \cdot y + 2y \cdot z + 7z \cdot y + 7z \cdot z\)
= \(2y^2 + 2yz + 7yz + 7z^2\)
Step 3: Combine both results:
= \(9y^2 – 16z^2 + 2y^2 + 9yz + 7z^2\)
Step 4: Final simplification:
= \(11y^2 + 9yz – 9z^2\)
Answer: 11y² + 9yz – 9z²

Q5: The adjacent sides of a rectangle are \(x^2-4xy+7y^2\) and \(x^3-5xy^2\). Find its area.

Step 1: Area of a rectangle = length × breadth
So we multiply the two given expressions: \[ (x^2 – 4xy + 7y^2)(x^3 – 5xy^2) \]Step 2: Expand using distributive property (each term of the first bracket multiplies each term of the second bracket): \[ = x^2(x^3 – 5xy^2) – 4xy(x^3 – 5xy^2) + 7y^2(x^3 – 5xy^2) \]Step 3: Multiply each term: \[ = x^5 – 5x^3y^2 \quad \text{(First term)} \\ – 4x^4y + 20x^2y^3 \quad \text{(Second term)} \\ + 7x^3y^2 – 35xy^4 \quad \text{(Third term)} \\ \]Step 4: Combine like terms: \[ = x^5 – 4x^4y + (-5x^3y^2 + 7x^3y^2) + 20x^2y^3 – 35xy^4 \\ = x^5 – 4x^4y + 2x^3y^2 + 20x^2y^3 – 35xy^4 \]Answer: Area of the rectangle = \(x^5 – 4x^4y + 2x^3y^2 + 20x^2y^3 – 35xy^4\)

Q6: The base and the altitude of a triangle are \(\left(3x-4y\right)\) and \(\left(6x+5y\right)\) respectively. Find its area.

Step 1: Area of a triangle is given by the formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Substitute the given expressions: \[ \text{Area} = \frac{1}{2} \times (3x – 4y) \times (6x + 5y) \]Step 2: Expand the binomials using distributive property: \[ (3x – 4y)(6x + 5y) = 3x \cdot 6x + 3x \cdot 5y – 4y \cdot 6x – 4y \cdot 5y \\ = 18x^2 + 15xy – 24xy – 20y^2 \]Step 3: Combine like terms: \[ = 18x^2 – 9xy – 20y^2 \]Step 4: Multiply by \(\frac{1}{2}\): \[ \text{Area} = \frac{1}{2}(18x^2 – 9xy – 20y^2) = 9x^2 – \frac{9}{2}xy – 10y^2 \]Answer: Area of the triangle = \(9x^2 – \frac{9}{2}xy – 10y^2\)

Q7: Multiply \(-4xy^3\) and \(6x^2y\) and verify your result for x = 2 and y = 1.

i. Multiplying the expressions
Step 1: Multiply the coefficients and the variables: \[ (-4xy^3) \cdot (6x^2y) = (-4 \cdot 6)(x \cdot x^2)(y^3 \cdot y) \]Step 2: Simplify: \[ = -24x^3y^4 \]ii. Verification for \(x = 2\), \(y = 1\)
Step 3: Evaluate both original expressions:
First term: \[ -4xy^3 = -4(2)(1)^3 = -8 \] Second term: \[ 6x^2y = 6(2)^2(1) = 6(4)(1) = 24 \]Step 4: Multiply the evaluated values: \[ -8 \times 24 = -192 \]Step 5: Now verify the simplified expression \(-24x^3y^4\):
Substitute \(x = 2\), \(y = 1\): \[ -24(2)^3(1)^4 = -24(8)(1) = -192 \]Answer: Product = \(-24x^3y^4\); Verified for \(x = 2\), \(y = 1\): Result = \(-192\)

Q8: Find the value of \(\left(3x^3\right)\times\left(-5xy^2\right)\times\left(2x^2yz^3\right)\) for x = 1, y = 2 and z = 3.

i. Multiply the expressions algebraically
Step 1: Multiply the coefficients and like terms: \[ (3x^3) \cdot (-5xy^2) \cdot (2x^2yz^3) \\ = (3 \cdot -5 \cdot 2)(x^3 \cdot x \cdot x^2)(y^2 \cdot y)(z^3) \]Step 2: Simplify: \[ = -30x^6y^3z^3 \]ii. Substitute \(x = 1\), \(y = 2\), \(z = 3\) into the expression
Step 3: Evaluate: \[ -30(1)^6(2)^3(3)^3 = -30(1)(8)(27) \]Step 4: Final calculation: \[ -30 \times 8 = -240,\quad -240 \times 27 = -6480 \]Answer: Value = \(-6480\)

Q9: Evaluate \(\left(3x^4y^2\right)\left(2x^2y^3\right)\) for x = 1 and y = 2.

i. Multiply the algebraic expressions
Step 1: Multiply the coefficients and variables: \[ (3x^4y^2) \times (2x^2y^3) = (3 \times 2)(x^{4+2})(y^{2+3}) = 6x^6y^5 \]ii. Substitute values \(x=1\), \(y=2\)
Step 2: Calculate powers: \[ x^6 = 1^6 = 1, \quad y^5 = 2^5 = 32 \]Step 3: Substitute and simplify: \[ 6 \times 1 \times 32 = 192 \]Answer: Value of the expression = \(192\)

Q10: Evaluate \(\left(x^5\right)\times\left(3x^2\right)\times\left(-2x\right)\) for x = 1

i. Multiply the expressions algebraically
Step 1: Multiply coefficients and powers of \(x\): \[ x^5 \times 3x^2 \times (-2x) = (1 \times 3 \times -2) \times x^{5 + 2 + 1} = -6x^8 \]ii. Substitute \(x = 1\)
Step 2: Calculate: \[ -6 \times (1)^8 = -6 \times 1 = -6 \]Answer: Value of the expression = \(-6\)

Q11: Divide:

i. \(-70a^3\) by \(14a^2\)

Step 1: Divide coefficients: \(\frac{-70}{14} = -5\)
Step 2: Subtract powers of \(a\): \(a^{3-2} = a^1 = a\)
Answer: \(-5a\)

ii. \(24x^3y^3\) by \(-8y^2\)

Step 1: Divide coefficients: \(\frac{24}{-8} = -3\)
Step 2: Divide variables: \(x^3 \div 1 = x^3\), \(y^{3-2} = y^1 = y\)
Answer: \(-3x^3y\)

iii. \(15a^4b\) by \(-5a^3b\)

Step 1: Divide coefficients: \(\frac{15}{-5} = -3\)
Step 2: Divide variables: \(a^{4-3} = a^1 = a\), \(b^{1-1} = b^0 = 1\)
Answer: \(-3a\)

iv. \(-24x^4d^3\) by \(-2x^2d^5\)

Step 1: Divide coefficients: \(\frac{-24}{-2} = 12\)
Step 2: Divide variables: \(x^{4-2} = x^2\), \(d^{3-5} = d^{-2} = \frac{1}{d^2}\)
Answer: \(12x^2 \div d^2 = \frac{12x^2}{d^2}\)

v. \(63a^4b^5c^6\) by \(-9a^2b^4c^3\)

Step 1: Divide coefficients: \(\frac{63}{-9} = -7\)
Step 2: Divide variables: \(a^{4-2} = a^2\), \(b^{5-4} = b^1 = b\), \(c^{6-3} = c^3\)
Answer: \(-7a^2bc^3\)

vi. \(8x – 10y + 6c\) by \(2\)

Step 1: Divide each term by 2: \[ \frac{8x}{2} – \frac{10y}{2} + \frac{6c}{2} = 4x – 5y + 3c \] Answer: \(4x – 5y + 3c\)

vii. \(15a^3b^4 – 10a^4b^3 – 25a^3b^6\) by \(-5a^3b^2\)

Step 1: Divide each term separately: \[ \frac{15a^3b^4}{-5a^3b^2} = -3b^{4-2} = -3b^2 \\ \frac{-10a^4b^3}{-5a^3b^2} = 2a^{4-3}b^{3-2} = 2ab \\ \frac{-25a^3b^6}{-5a^3b^2} = 5b^{6-2} = 5b^4 \] Answer: \(-3b^2 + 2ab + 5b^4\)

viii. \(-14x^6y^3 – 21x^4y^5 + 7x^5y^4\) by \(7x^2y^2\)

Step 1: Divide each term: \[ \frac{-14x^6y^3}{7x^2y^2} = -2x^{6-2}y^{3-2} = -2x^4y \\ \frac{-21x^4y^5}{7x^2y^2} = -3x^{4-2}y^{5-2} = -3x^2y^3 \\ \frac{7x^5y^4}{7x^2y^2} = x^{5-2}y^{4-2} = x^3y^2 \] Answer: \(-2x^4y – 3x^2y^3 + x^3y^2\)

ix. \(a^2 + 7a + 12\) by \(a + 4\)

Step 1: Perform polynomial division:
Divide \(a^2\) by \(a\): \(a\)
Multiply \(a\) by \(a + 4\): \(a^2 + 4a\)
Subtract: \((a^2 + 7a + 12) – (a^2 + 4a) = 3a + 12\)
Divide \(3a\) by \(a\): \(3\)
Multiply \(3\) by \(a + 4\): \(3a + 12\)
Subtract: \((3a + 12) – (3a + 12) = 0\)
Answer: \(a + 3\)

x. \(x^2 + 3x – 54\) by \(x – 6\)

Step 1: Polynomial division:
Divide \(x^2\) by \(x\): \(x\)
Multiply \(x\) by \(x – 6\): \(x^2 – 6x\)
Subtract: \((x^2 + 3x – 54) – (x^2 – 6x) = 9x – 54\)
Divide \(9x\) by \(x\): \(9\)
Multiply \(9\) by \(x – 6\): \(9x – 54\)
Subtract: \((9x – 54) – (9x – 54) = 0\)
Answer: \(x + 9\)

xi. \(12x^2 + 7xy – 12y^2\) by \(3x + 4y\)

Step 1: Polynomial division:
Divide \(12x^2\) by \(3x\): \(4x\)
Multiply \(4x\) by \(3x + 4y\): \(12x^2 + 16xy\)
Subtract: \((12x^2 + 7xy – 12y^2) – (12x^2 + 16xy) = -9xy – 12y^2\)
Divide \(-9xy\) by \(3x\): \(-3y\)
Multiply \(-3y\) by \(3x + 4y\): \(-9xy – 12y^2\)
Subtract: \((-9xy – 12y^2) – (-9xy – 12y^2) = 0\)
Answer: \(4x – 3y\)

xii. \(x^6 – 8\) by \(x^2 – 2\)

Step 1: Polynomial division:
Divide \(x^6\) by \(x^2\): \(x^4\)
Multiply \(x^4\) by \(x^2 – 2\): \(x^6 – 2x^4\)
Subtract: \((x^6 – 8) – (x^6 – 2x^4) = 2x^4 – 8\)
Divide \(2x^4\) by \(x^2\): \(2x^2\)
Multiply \(2x^2\) by \(x^2 – 2\): \(2x^4 – 4x^2\)
Subtract: \((2x^4 – 8) – (2x^4 – 4x^2) = 4x^2 – 8\)
Divide \(4x^2\) by \(x^2\): \(4\)
Multiply \(4\) by \(x^2 – 2\): \(4x^2 – 8\)
Subtract: \((4x^2 – 8) – (4x^2 – 8) = 0\)
Answer: \(x^4 + 2x^2 + 4\)

xiii. \(6x^3 – 13x^2 – 13x + 30\) by \(2x^2 – x – 6\)

Step 1: Polynomial division:
Divide \(6x^3\) by \(2x^2\): \(3x\)
Multiply \(3x\) by \(2x^2 – x – 6\): \(6x^3 – 3x^2 – 18x\)
Subtract: \((6x^3 – 13x^2 – 13x + 30) – (6x^3 – 3x^2 – 18x) = -10x^2 + 5x + 30\)
Divide \(-10x^2\) by \(2x^2\): \(-5\)
Multiply \(-5\) by \(2x^2 – x – 6\): \(-10x^2 + 5x + 30\)
Subtract: \((-10x^2 + 5x + 30) – (-10x^2 + 5x + 30) = 0\)
Answer: \(3x – 5\)

xiv. \(4a^2 + 12ab + 9b^2 – 25c^2\) by \(2a + 3b + 5c\)

Step 1: Polynomial division:
Divide \(4a^2\) by \(2a\): \(2a\)
Multiply \(2a\) by \(2a + 3b + 5c\): \(4a^2 + 6ab + 10ac\)
Subtract: \((4a^2 + 12ab + 9b^2 – 25c^2) – (4a^2 + 6ab + 10ac) = 6ab + 9b^2 – 10ac – 25c^2\)
Divide \(6ab\) by \(2a\): \(3b\)
Multiply \(3b\) by \(2a + 3b + 5c\): \(6ab + 9b^2 + 15bc\)
Subtract: \((6ab + 9b^2 – 10ac – 25c^2) – (6ab + 9b^2 + 15bc) = -10ac – 15bc – 25c^2\)
Divide \(-10ac\) by \(2a\): \(-5c\)
Multiply \(-5c\) by \(2a + 3b + 5c\): \(-10ac – 15bc – 25c^2\)
Subtract: \((-10ac – 15bc – 25c^2) – (-10ac – 15bc – 25c^2) = 0\)
Answer: \(2a + 3b – 5c\)

xv. \(16 + 8x + x^6 – 8x^3 – 2x^4 + x^2\) by \(x + 4 – x^3\)

Step 1: Rewrite divisor: \(x + 4 – x^3 = -x^3 + x + 4\)
Polynomial division is complex; rearranging dividend and divisor in descending powers:
Dividend: \(x^6 – 8x^3 – 2x^4 + x^2 + 8x + 16\)
Sort dividend: \(x^6 – 2x^4 – 8x^3 + x^2 + 8x + 16\)
Divisor: \(-x^3 + x + 4\)
Proceed polynomial division:
Divide \(x^6\) by \(-x^3\): \(-x^3\)
Multiply \(-x^3\) by \(-x^3 + x + 4\): \(x^6 – x^4 – 4x^3\)
Subtract: \[ (x^6 – 2x^4 – 8x^3 + x^2 + 8x + 16) – (x^6 – x^4 – 4x^3) = (-2x^4 + x^4) + (-8x^3 + 4x^3) + x^2 + 8x + 16 \\ = -x^4 – 4x^3 + x^2 + 8x + 16 \] Divide \(-x^4\) by \(-x^3\): \(x\)
Multiply \(x\) by \(-x^3 + x + 4\): \(-x^4 + x^2 + 4x\)
Subtract: \[ (-x^4 – 4x^3 + x^2 + 8x + 16) – (-x^4 + x^2 + 4x) = -4x^3 + (x^2 – x^2) + (8x – 4x) + 16 \\ = -4x^3 + 4x + 16 \] Divide \(-4x^3\) by \(-x^3\): 4
Multiply \(4\) by \(-x^3 + x + 4\): \(-4x^3 + 4x + 16\)
Subtract: \[ (-4x^3 + 4x + 16) – (-4x^3 + 4x + 16) = 0 \] Answer: \(-x^3 + x + 4\)

Q12: Find the quotient and the remainder (if any), when:

i. \(a^3 – 5a^2 + 8a + 15\) is divided by \(a + 1\)

Step 1: Divide \(a^3\) by \(a\): \(a^2\)
Step 2: Multiply \(a^2\) by divisor \(a+1\): \(a^3 + a^2\)
Step 3: Subtract:
\((a^3 – 5a^2 + 8a + 15) – (a^3 + a^2) = -6a^2 + 8a + 15\)
Step 4: Divide \(-6a^2\) by \(a\): \(-6a\)
Step 5: Multiply \(-6a\) by \(a+1\): \(-6a^2 – 6a\)
Step 6: Subtract:
\((-6a^2 + 8a + 15) – (-6a^2 – 6a) = 14a + 15\)
Step 7: Divide \(14a\) by \(a\): \(14\)
Step 8: Multiply \(14\) by \(a+1\): \(14a + 14\)
Step 9: Subtract:
\((14a + 15) – (14a + 14) = 1\)
Step 10: Quotient = \(a^2 – 6a + 14\), remainder = 1
Verification: \[ \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ = (a + 1)(a^2 – 6a + 14) + 1 \\ = a^3 – 6a^2 + 14a + a^2 – 6a + 14 + 1 = a^3 – 5a^2 + 8a + 15 \] Verified!
Answer: Quotient = \(a^2 – 6a + 14\), Remainder = 1

ii. \(3x^4 + 6x^3 – 6x^2 + 2x – 7\) is divided by \(x – 3\)

Step 1: Divide \(3x^4\) by \(x\): \(3x^3\)
Step 2: Multiply \(3x^3\) by divisor \(x – 3\): \(3x^4 – 9x^3\)
Step 3: Subtract:
\((3x^4 + 6x^3 – 6x^2 + 2x – 7) – (3x^4 – 9x^3) = 15x^3 – 6x^2 + 2x – 7\)
Step 4: Divide \(15x^3\) by \(x\): \(15x^2\)
Step 5: Multiply \(15x^2\) by \(x – 3\): \(15x^3 – 45x^2\)
Step 6: Subtract:
\((15x^3 – 6x^2 + 2x – 7) – (15x^3 – 45x^2) = 39x^2 + 2x – 7\)
Step 7: Divide \(39x^2\) by \(x\): \(39x\)
Step 8: Multiply \(39x\) by \(x – 3\): \(39x^2 – 117x\)
Step 9: Subtract:
\((39x^2 + 2x – 7) – (39x^2 – 117x) = 119x – 7\)
Step 10: Divide \(119x\) by \(x\): \(119\)
Step 11: Multiply \(119\) by \(x – 3\): \(119x – 357\)
Step 12: Subtract:
\((119x – 7) – (119x – 357) = 350\)
Step 13: Quotient = \(3x^3 + 15x^2 + 39x + 119\), remainder = 350
Verification: \[ \text{Dividend} = (x – 3)(3x^3 + 15x^2 + 39x + 119) + 350 \] Multiply and verify the expression equals original dividend (left as exercise).
Answer: Quotient = \(3x^3 + 15x^2 + 39x + 119\), Remainder = 350

iii. \(6x^2 + x – 15\) is divided by \(3x + 5\)

Step 1: Divide \(6x^2\) by \(3x\): \(2x\)
Step 2: Multiply \(2x\) by divisor \(3x + 5\): \(6x^2 + 10x\)
Step 3: Subtract:
\((6x^2 + x – 15) – (6x^2 + 10x) = -9x – 15\)
Step 4: Divide \(-9x\) by \(3x\): \(-3\)
Step 5: Multiply \(-3\) by \(3x + 5\): \(-9x – 15\)
Step 6: Subtract:
\((-9x – 15) – (-9x – 15) = 0\)
Step 7: Quotient = \(2x – 3\), remainder = 0
Verification: \[ \text{Dividend} = (3x + 5)(2x – 3) \\ = 6x^2 – 9x + 10x – 15 = 6x^2 + x – 15 \] Verified!
Answer: Quotient = \(2x – 3\), Remainder = 0

Q13: The area of rectangle is \(x^3-8x^2+7\) and one of its sides is \(x-1\). Find the length of the adjacent side.

Area \(= x^3 – 8x^2 + 7\), one side \(= x – 1\)
Step 1: Let the adjacent side be \(L\). Since area = length × breadth, we have: \[ (x – 1) \times L = x^3 – 8x^2 + 7 \]Step 2: To find \(L\), divide the area by the given side: \[ L = \frac{x^3 – 8x^2 + 7}{x – 1} \]Step 3: Perform polynomial division of \(x^3 – 8x^2 + 7\) by \(x – 1\):
Divide \(x^3\) by \(x\): \(x^2\)
Multiply \(x^2 \times (x – 1) = x^3 – x^2\)
Subtract: \((x^3 – 8x^2 + 7) – (x^3 – x^2) = -7x^2 + 7\)
Divide \(-7x^2\) by \(x\): \(-7x\)
Multiply \(-7x \times (x – 1) = -7x^2 + 7x\)
Subtract: \((-7x^2 + 7) – (-7x^2 + 7x) = -7x + 7\)
Divide \(-7x\) by \(x\): \(-7\)
Multiply \(-7 \times (x – 1) = -7x + 7\)
Subtract: \((-7x + 7) – (-7x + 7) = 0\)
Step 4: Quotient is \(x^2 – 7x – 7\) and remainder is 0.
Answer: Length of the adjacent side = \(x^2 – 7x – 7\)