Fill in the blanks:
(i) 6×3 = ............... and 6x×3x = ...............
(ii) 6×3 = ............... and 6x2×3x3 = ...............
(iii) 5×4 = ............... and 5x×4y = ...............
(iv) 4×7 = ............... and 4ax×7x = ...............
(v) 6×2 = ............... and 6xy×2xy = ...............
(vi) 12×4 = ............... and 12ax2×4ax = ...............
(vii) 1×8 = ............... and a2xy2×8a3x2y = ...............
(viii) 15×3 = ............... and 15x×3x5y2 = ...............
Answer
The product of the given monomials = (product of their numeral coefficients) × (product of their literals), and in multiplication the powers of like factors are added.
(i) 6×3= 18 and 6x×3x=(6×3)(x×x)=18x2.
(ii) 6×3= 18 and 6x2×3x3=(6×3)(x2+3)=18x5.
(iii) 5×4= 20 and 5x×4y=(5×4)(x×y)=20xy.
(iv) 4×7= 28 and 4ax×7x=(4×7)(a×x1+1)=28ax2.
(v) 6×2= 12 and 6xy×2xy=(6×2)(x1+1×y1+1)=12x2y2.
(vi) 12×4= 48 and 12ax2×4ax=(12×4)(a1+1×x2+1)=48a2x3.
(vii) 1×8= 8 and a2xy2×8a3x2y=(1×8)(a2+3×x1+2×y2+1)=8a5x3y3.
(viii) 15×3= 45 and 15x×3x5y2=(15×3)(x1+5×y2)=45x6y2.
Fill in the blanks:
(i) 4x×6x×2 = ...............
(ii) 3ab×6ax = ...............
(iii) x×2x2×3x3 = ...............
(iv) 5×5a3 = ...............
(v) 6×6x2×6x2y2 = ...............
(vi) −8x×−3x = ...............
(vii) −5×−3x×5x2 = ...............
(viii) 8×−4xy2×3x3y2 = ...............
(ix) −4x×5xy×3z = ...............
(x) 5x×2x2y×−7y3×2x3y2 = ...............
Answer
The product of the given monomials = (product of their numeral coefficients) × (product of their literals), and in multiplication the powers of like factors are added.
(i) 4x×6x×2=(4×6×2)(x1+1)=48x2.
(ii) 3ab×6ax=(3×6)(a1+1×b×x)=18a2bx.
(iii) x×2x2×3x3=(1×2×3)(x1+2+3)=6x6.
(iv) 5×5a3=(5×5)a3=25a3.
(v) 6×6x2×6x2y2=(6×6×6)(x2+2×y2)=216x4y2.
(vi) −8x×−3x=(−8×−3)(x1+1)=24x2.
(vii) −5×−3x×5x2=(−5×−3×5)(x1+2)=75x3.
(viii) 8×−4xy2×3x3y2=(8×−4×3)(x1+3×y2+2)=−96x4y4.
(ix) −4x×5xy×3z=(−4×5×3)(x1+1×y×z)=−60x2yz.
(x) 5x×2x2y×−7y3×2x3y2=(5×2×−7×2)(x1+2+3×y1+3+2)=−140x6y6.
Find the value of:
3x3×5x4
Answer
Solving,
⇒3x3×5x4⇒(3×5)(x3+4)⇒15x7
Hence, 3x3×5x4=15x7.
Find the value of:
5a2×7a7
Answer
Solving,
⇒5a2×7a7⇒(5×7)(a2+7)⇒35a9
Hence, 5a2×7a7=35a9.
Find the value of:
3abc×6ac3
Answer
Solving,
⇒3abc×6ac3⇒(3×6)(a1+1×b×c1+3)⇒18a2bc4
Hence, 3abc×6ac3=18a2bc4.
Find the value of:
a2b2×5a3b4
Answer
Solving,
⇒a2b2×5a3b4⇒(1×5)(a2+3×b2+4)⇒5a5b6
Hence, a2b2×5a3b4=5a5b6.
Find the value of:
2x2y3×5x3y4
Answer
Solving,
⇒2x2y3×5x3y4⇒(2×5)(x2+3×y3+4)⇒10x5y7
Hence, 2x2y3×5x3y4=10x5y7.
Find the value of:
abc×bcd
Answer
Solving,
⇒abc×bcd⇒a×b1+1×c1+1×d⇒ab2c2d
Hence, abc×bcd=ab2c2d.
Multiply:
a+b by ab
Answer
Solving,
×aa2b++babab2
Hence, (a+b)×ab=a2b+ab2.
Multiply:
3ab−4b by 3ab
Answer
Solving,
×3ab9a2b2−−4b3ab12ab2
Hence, (3ab−4b)×3ab=9a2b2−12ab2.
Multiply:
2xy−5by by 4bx
Answer
Solving,
×2xy8bx2y−−5by4bx20b2xy
Hence, (2xy−5by)×4bx=8bx2y−20b2xy.
Multiply:
4x+2y by 3xy
Answer
Solving,
×4x12x2y++2y3xy6xy2
Hence, (4x+2y)×3xy=12x2y+6xy2.
Multiply:
x2−x by 2x
Answer
Solving,
×x22x3−−x2x2x2
Hence, (x2−x)×2x=2x3−2x2.
Multiply:
1+4x by x
Answer
Solving,
×1x++4xx4x2
Hence, (1+4x)×x=x+4x2.
Multiply:
9xy2+3x2y by 5xy
Answer
Solving,
×9xy245x2y3++3x2y5xy15x3y2
Hence, (9xy2+3x2y)×5xy=45x2y3+15x3y2.
Multiply:
6x−5y by 3axy
Answer
Solving,
×6x18ax2y−−5y3axy15axy2
Hence, (6x−5y)×3axy=18ax2y−15axy2.
Multiply:
−x+y−z and −2x
Answer
Solving,
×−x2x2+−y2xy−+z−2x2xz
Hence, (−x+y−z)×(−2x)=2x2−2xy+2xz.
Multiply:
xy−yz and x2yz2
Answer
Solving,
×xyx3y2z2−−yzx2yz2x2y2z3
Hence, (xy−yz)×x2yz2=x3y2z2−x2y2z3.
Multiply:
2xyz+3xy and −2y2z
Answer
Solving,
×2xyz−4xy3z2+−3xy−2y2z6xy3z
Hence, (2xyz+3xy)×(−2y2z)=−4xy3z2−6xy3z.
Multiply:
−3xy2+4x2y and −xy
Answer
Solving,
×−3xy23x2y3+−4x2y−xy4x3y2
Hence, (−3xy2+4x2y)×(−xy)=3x2y3−4x3y2.
Multiply:
4xy and −x2y−3x2y2
Answer
Solving,
×−x2y−4x3y2−−3x2y24xy12x3y3
Hence, 4xy×(−x2y−3x2y2)=−4x3y2−12x3y3.
Multiply:
3a+4b−5c and 3a
Answer
Solving,
×3a9a2++4b12ab−−5c3a15ac
Hence, (3a+4b−5c)×3a=9a2+12ab−15ac.
Multiply:
−5xy and −xy2−6x2y
Answer
Solving,
×−xy25x2y3−+6x2y−5xy30x3y2
Hence, −5xy×(−xy2−6x2y)=5x2y3+30x3y2.
Multiply:
x+2 and x+10
Answer
Solving,
⇒(x+2)(x+10)⇒x2+10x+2x+20⇒x2+12x+20.
Hence, (x+2)(x+10)=x2+12x+20.
Multiply:
x+5 and x−3
Answer
Solving,
⇒(x+5)(x−3)⇒x2−3x+5x−15⇒x2+2x−15.
Hence, (x+5)(x−3)=x2+2x−15.
Multiply:
x−5 and x+3
Answer
Solving,
⇒(x−5)(x+3)⇒x2+3x−5x−15⇒x2−2x−15.
Hence, (x−5)(x+3)=x2−2x−15.
Multiply:
x−5 and x−3
Answer
Solving,
⇒(x−5)(x−3)⇒x2−3x−5x+15⇒x2−8x+15.
Hence, (x−5)(x−3)=x2−8x+15.
Multiply:
2x+y and x+3y
Answer
Solving,
⇒(2x+y)(x+3y)⇒2x2+6xy+xy+3y2⇒2x2+7xy+3y2.
Hence, (2x+y)(x+3y)=2x2+7xy+3y2.
Multiply:
3x−5y and x+6y
Answer
Solving,
⇒(3x−5y)(x+6y)⇒3x2+18xy−5xy−30y2⇒3x2+13xy−30y2.
Hence, (3x−5y)(x+6y)=3x2+13xy−30y2.
Multiply:
x+9y and x−5y
Answer
Solving,
⇒(x+9y)(x−5y)⇒x2−5xy+9xy−45y2⇒x2+4xy−45y2.
Hence, (x+9y)(x−5y)=x2+4xy−45y2.
Multiply:
2x+5y and 2x+5y
Answer
Solving,
⇒(2x+5y)(2x+5y)⇒4x2+10xy+10xy+25y2⇒4x2+20xy+25y2.
Hence, (2x+5y)(2x+5y)=4x2+20xy+25y2.
Multiply:
3abc and −5a2b2c
Answer
Solving,
×3abc−5a2b2c−15a3b3c2
Hence, 3abc×(−5a2b2c)=−15a3b3c2.
Multiply:
x−y+z and −2x
Answer
Solving,
×x−2x2−+y2xy+−z−2x2xz
Hence, (x−y+z)×(−2x)=−2x2+2xy−2xz.
Multiply:
2x−3y−5z and −2y
Answer
Solving,
×2x−4xy−+3y6y2−+5z−2y10yz
Hence, (2x−3y−5z)×(−2y)=−4xy+6y2+10yz.
Multiply:
−8xyz+10x2yz3 and xyz
Answer
Solving,
×−8xyz−8x2y2z2++10x2yz3xyz10x3y2z4
Hence, (−8xyz+10x2yz3)×xyz=−8x2y2z2+10x3y2z4.
Multiply:
xyz and −13xy2z+15x2yz−6xyz2
Answer
Solving,
×−13xy2z−13x2y3z2++15x2yz15x3y2z2−−6xyz2xyz6x2y2z3
Hence, xyz×(−13xy2z+15x2yz−6xyz2)=−13x2y3z2+15x3y2z2−6x2y2z3.
Multiply:
4abc−5a2bc−6ab2c and −2abc2
Answer
Solving,
×4abc−8a2b2c3−+5a2bc10a3b2c3−+6ab2c−2abc212a2b3c3
Hence, (4abc−5a2bc−6ab2c)×(−2abc2)=−8a2b2c3+10a3b2c3+12a2b3c3.
Find the product of:
xy−ab and xy+ab
Answer
Solving,
⇒(xy−ab)(xy+ab)⇒xy×xy+xy×ab−ab×xy−ab×ab⇒x2y2+abxy−abxy−a2b2⇒x2y2−a2b2.
Hence, (xy−ab)(xy+ab)=x2y2−a2b2.
Find the product of:
2abc−3xy and 2abc+3xy
Answer
Solving,
⇒(2abc−3xy)(2abc+3xy)⇒2abc×2abc+2abc×3xy−3xy×2abc−3xy×3xy⇒4a2b2c2+6abcxy−6abcxy−9x2y2⇒4a2b2c2−9x2y2.
Hence, (2abc−3xy)(2abc+3xy)=4a2b2c2−9x2y2.
Find the product of:
a+b−c and 2a−3b
Answer
Solving,
⇒(a+b−c)(2a−3b)⇒a(2a−3b)+b(2a−3b)−c(2a−3b)⇒2a2−3ab+2ab−3b2−2ac+3bc⇒2a2−ab−2ac−3b2+3bc.
Hence, (a+b−c)(2a−3b)=2a2−ab−2ac−3b2+3bc.
Find the product of:
5x−6y−7z and 2x+3y
Answer
Solving,
⇒(5x−6y−7z)(2x+3y)⇒5x(2x+3y)−6y(2x+3y)−7z(2x+3y)⇒10x2+15xy−12xy−18y2−14xz−21yz⇒10x2+3xy−14xz−18y2−21yz.
Hence, (5x−6y−7z)(2x+3y)=10x2+3xy−14xz−18y2−21yz.
Find the product of:
5x−6y−7z and 2x+3y+z
Answer
Solving,
⇒(5x−6y−7z)(2x+3y+z)⇒5x(2x+3y+z)−6y(2x+3y+z)−7z(2x+3y+z)⇒10x2+15xy+5zx−12xy−18y2−6yz−14zx−21yz−7z2⇒10x2+3xy−9zx−18y2−27yz−7z2.
Hence, (5x−6y−7z)(2x+3y+z)=10x2+3xy−9zx−18y2−27yz−7z2.
Find the product of:
2a+3b−4c and a−b−c
Answer
Solving,
⇒(2a+3b−4c)(a−b−c)⇒2a(a−b−c)+3b(a−b−c)−4c(a−b−c)⇒2a2−2ab−2ac+3ab−3b2−3bc−4ac+4bc+4c2⇒2a2+ab−6ac−3b2+bc+4c2.
Hence, (2a+3b−4c)(a−b−c)=2a2+ab−6ac−3b2+bc+4c2.