Class 8: Fundamental Concepts and Operations on Algebraic Expressions – Exercise 16C – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Multiply:

$\displaystyle \text{ i) } 8x^2 \text{ by } -5x$

$\displaystyle \text{ ii) } \frac{2}{3} ab \text{ by } -6a^2b$

$\displaystyle \text{ iii) } 7x^2y^3 \text{ by } -4x^3y$

$\displaystyle \text{iv) } \Big( \frac{-5}{9} ax^2 \Big) \text{ by } \Big ( \frac{-3}{5} bxa^2 \Big)$

$\displaystyle \text{v) } \Big( \frac{-2}{3} a^6b^5 \Big) \text{ by } \Big ( \frac{-9}{4} a^3b^3 \Big)$

$\displaystyle \text{vi) } \ \Big( \frac{-5}{8} p^2q \Big) \text{ by } \Big ( \frac{16}{25} {pq}^2 \Big)$

Answer:

$\displaystyle \text{ i) } {8x}^2 \times ( -5x ) = -40x^3$

$\displaystyle \text{ ii) } \frac{2}{3} ab \times ( -6a^2b ) = -{4a}^3b^2$

$\displaystyle \text{ iii) } {7x}^2y^3 \times ( -4x^3y ) = -28x^5y^4$

i $\displaystyle \text{v) } \Big( \frac{-5}{9} {ax}^2 \Big) \times \Big( \frac{-3}{5} {bxa}^2 \Big) = \frac{1}{3} x^3 a^3 b$

$\displaystyle \text{v) } \Big( \frac{-2}{3} a^6b^5 \Big) \times \Big( \frac{-9}{4} a^3b^3 \Big) = \frac{3}{2} a^9b^8$

$\displaystyle \text{vi) } \Big( \frac{-5}{8} p^2q \Big) \times \Big( \frac{16}{25} {pq}^2 \Big) = \frac{-2}{5} p^3q^3$

$\displaystyle \\$

Question 2: Multiply:

$\displaystyle \text{ i) } ( 2x^2-5x+6 ) \text{ by } -3x$

$\displaystyle \text{ ii) } ( 3-2y-y^2 ) \text{ by } -4xy$

$\displaystyle \text{ iii) } ( {3x}^2y-2xy^2+5xy-6 ) \text{ by } 4xy$

i $\displaystyle \text{v) } ( 7a^3-5ab^2-{2b}^3+3ab+2a-5 ) \text{ by } -3{ab}^2$

Answer:

$\displaystyle \text{ i) } ( 2x^2-5x+6 ) \times ( -3x ) = -6x^3+15x^2-18x$

$\displaystyle \text{ ii) } ( 3-2y-y^2 ) \times ( -4xy ) = -12xy+8{xy}^2+{4xy}^3$

$\displaystyle \text{ iii) } ( {3x}^2y-2xy^2+5xy-6 ) \times 4xy = {12x}^3y^2-{8x}^2y^3+20x^2y^2-24xy$

i $\displaystyle \text{v) } ( 7a^3-5ab^2-{2b}^3+3ab+2a-5 ) \times ( -3{ab}^2 )$

$\displaystyle = b^2+15\ a^2b^4+{6ab}^5-9\ a^2b^3-{6a}^2b^2+{15ab}^2$

$\displaystyle \\$

Question 3: Multiply:

$\displaystyle \text{ i) } ( a+5 ) \text{ by } ( a+4 )$ $\displaystyle \text{ ii) } ( x-3 ) \text{ by } ( x+8 )$

$\displaystyle \text{ iii) } ( y-4 ) \text{ by } ( y-6 )$ i $\displaystyle \text{v) } ( z+1 ) \text{ by } ( z-7 )$

$\displaystyle \text{v) } ( 3a-2b ) \text{ by } ( 2a+3b )$ $\displaystyle \text{vi) } ( 5x+4y ) \text{ by } ( 2x-3y )$

Answer:

$\displaystyle \text{ i) } ( a+5 ) ( a+4 ) = a^2+9a+20$

$\displaystyle \text{ ii) } ( x-3 ) ( x+8 ) = x^2-3x+8x-24 = x^2+5x-24$

$\displaystyle \text{ iii) } ( y-4 ) ( y-6 ) = y^2-4y-6y+24 = y^2+10y+24$

i $\displaystyle \text{v) } ( z+1 ) ( z-7 ) = \ z^2+z-7z-7 = z^2-6z-7$

$\displaystyle \text{v) } ( 3a-2b ) ( 2a+3b ) = {6a}^2-4ab+9ab-{6b}^2 = {6a}^2+5ab-{6b}^2$

$\displaystyle \text{vi) } ( 5x+4y ) ( 2x-3y ) = {10x}^2+8xy-15xy-12y^2 = 10x^2-7xy-12y^2$

$\displaystyle \\$

Question 4: Multiply:

$\displaystyle \text{ i) } (4x^2+3x-5) \text{ by } (2x+3)$ $\displaystyle \text{ ii) } (3-2x+5x^2) \text{ by } (5x-4)$

$\displaystyle \text{ iii) } ( x^3-5x+3 ) \text{ by } ( 2x+9 )$ i $\displaystyle \text{v) } ( 4x^2+xy+{9y}^2 ) \text{ by } ( 2x-3y )$

Answer:

$\displaystyle \text{ i) } (4x^2+3x-5) \times (2x+3)$

$\displaystyle = 8x^3+{6x}^2-10x+{12x}^2+9x-15$

$\displaystyle = {8x}^3+{18x}^2-x-15$

$\displaystyle \text{ ii) } (3-2x+5x^2) \times (5x-4)$

$\displaystyle = 15x-10x^2+25x^3-12+8x-20x^2$

$\displaystyle = 25x^3-30x^2+23x-12$

$\displaystyle \text{ iii) } ( x^3-5x+3 ) ( 2x+9 )$

$\displaystyle = {2x}^4-{10x}^2+6x+{9x}^3-45x+27$

$\displaystyle = 2x^4+{9x}^3-10x^2-39x+27$

$\displaystyle \text{iv) } ( 4x^2+xy+{9y}^2 ) \times ( 2x-3y )$

$\displaystyle = {8x}^3+{2x}^2y+18{xy}^2-12x^2y-{6xy}^2-27y^3$

$\displaystyle = {8x}^3-27y^3-10x^2y+12{xy}^2$

$\displaystyle \\$

Question 5: Multiply:

$\displaystyle \text{ i) } ( {4x}^2-4x+1 ) \text{ by } ({2x}^2+x-2)$

$\displaystyle \text{ ii) } ( {3x}^2+4x-5 ) \text{ by } ( {4x}^2-7x+2 )$

$\displaystyle \text{ iii) } ( {4x}^2+24xy+{3y}^2 ) \text{ by } ( x^2-xy+y^2 )$

$\displaystyle \text{iv) } ( {6x}^3-{5x}^2+4x+1 ) \text{ by } ( x^2+7x-1 )$

$\displaystyle \text{v) } ( 8x^4-{3x}^2+9x-8 ) \text{ by } ( 2x^2-5x+3 )$

$\displaystyle \text{vi) } ( 3x^5-{7x}^3+2x^2-x+4 ) \text{ by } ( x^3-{2x}^2+3x-1 )$

Answer:

$\displaystyle \text{ i) } ( {4x}^2-4x+1 ) \times ({2x}^2+x-2)$

$\displaystyle = {8x}^4-{8x}^3+{2x}^2+{4x}^3-{4x}^2+x-{8x}^2+8x-2$

$\displaystyle = {8x}^4-{4x}^3-{10x}^2+9x-2$

$\displaystyle \text{ ii) } ( {3x}^2+4x-5 ) \times ( {4x}^2-7x+2 )$

$\displaystyle = 12x^4+{16x}^3-20x^2-{21x}^3-{28x}^2+35x+{6x}^2+8x-10$

$\displaystyle = 12x^4-5x^3-42x^2+43x-10$

$\displaystyle \text{ iii) } ( {4x}^2+24xy+{3y}^2 ) ( x^2-xy+y^2 )$

$\displaystyle = {4x}^4+{24x}^3y+{3x}^2y^2-{4x}^3y-24x^2y^2-3xy^3 \\ +4x^2y^2+24xy^3+4y^4$

$\displaystyle = {4x}^4+{20x}^3y-{17x}^2y^2+{21xy}^3+3y^4$

$\displaystyle \text{iv) } ( {6x}^3-{5x}^2+4x+1 ) \times ( x^2+7x-1 )$

$\displaystyle = 6x^5-{5x}^4+{4x}^3+x^2+{42x}^4-35x^3+28x^2+7x-6x^3 \\ +5x^2-4x-1+3x-1$

$\displaystyle = 6x^5+37x^4-37x^3+34x^2+3x-1$

$\displaystyle \text{v) } ( 8x^4-{3x}^2+9x-8 ) \times ( 2x^2-5x+3 )$

$\displaystyle = 16x^6-{6x}^4+{18x}^3-{16x}^2-40x^5 \\ +{15x}^3-45x^2+40x+{24x}^4-{9x}^2+27x-24$

$\displaystyle = 16x^6-{40x}^5+18x^4+33x^3-70x^2+67x-24$

$\displaystyle \text{vi) } ( 3x^5-{7x}^3+2x^2-x+4 ) \times ( x^3-{2x}^2+3x-1 )$

$\displaystyle = {3x}^8-7x^6+{2x}^5-x^4+{4x}^3+14x^5-4x^4 \\ +{2x}^3-6x^7-8x^2+9x^6-21x^4+{6x}^3-{3x}^2\\ +12x-3x^5 +{7x}^3-{2x}^2+x-4$

$\displaystyle = 3x^8+2x^6+13x^5-{26x}^4+{19x}^3-{6x}^7+{13x}^2+13x-4$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 8: Fundamental Concepts and Operations on Algebraic Expressions – Exercise 16C

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