What must be added to 2x4 5x3+ 2x2 x 3 so that the result is exactly divisible by x 2?
Solution
ANSWER:
Let k be added to 2x4–5x3+2x2–x–3 so that the result is exactly divisible by ( x – 2). Here, k is a constant.
∴
f(x)=2x4–5x3+2x2–x–3+k is exactly divisible by ( x – 2).
Using factor theorem, we have
f(2)=0
⇒2×24−5×23+2×22−2−3+k=0⇒32−40+8−5+k=0⇒−5+k=0⇒k=5
Thus, 5 must be added to
2x4–5x3+2x2–x–3 so that the result is exactly
divisible by ( x – 2).
- Textbook Solutions
- Class 8
- Math
- division of algebraic expressions
Rd Sharma Solutions for Class 8 Math Chapter 8 Division Of Algebraic Expressions are provided here with simple step-by-step explanations. These solutions for Division Of Algebraic Expressions are extremely popular among Class 8 students for Math Division Of Algebraic Expressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Book of Class 8 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Solutions. All Rd Sharma Solutions for class Class 8 Math are prepared by experts and are 100% accurate.
Page No 8.11:
Question 1:
Divide 5x3 − 15x2 + 25x by 5x.
Answer:
Page No 8.11:
Question 2:
Divide 4z3 + 6z2 − z by − 12z.
Answer:
Page No 8.11:
Question 3:
Divide 9x2y − 6xy + 12xy2 by −32xy.
Answer:
Page No 8.11:
Question 4:
Divide 3x3y2 + 2x2y + 15xy by 3xy.
Answer:
Page No 8.11:
Question 5:
Divide x2 + 7x + 12 by x + 4.
Answer:
Page No 8.11:
Question 6:
Divide 4y2 + 3y + 12 by 2y + 1.
Answer:
Page No 8.11:
Question 7:
Divide 3x3 + 4x2 + 5x + 18 by x + 2.
Answer:
Page No 8.11:
Question 8:
Divide 14x2 − 53x + 45 by 7x − 9.
Answer:
Page No 8.11:
Question 9:
Divide −21 + 71x − 31x2 − 24x3 by 3 − 8x.
Answer:
Page No 8.11:
Question 10:
Divide 3y4 − 3y3 − 4y2 − 4y by y2 − 2y.
Answer:
Page No 8.11:
Question 11:
Divide 2y5 + 10y4 + 6y3 + y2 + 5y + 3 by 2y3 + 1.
Answer:
Page No 8.11:
Question 12:
Divide x4 − 2x3 + 2x2 + x + 4 by x2 + x + 1.
Answer:
Page No 8.11:
Question 13:
Divide m3 − 14m2 + 37m − 26 by m2 − 12m +13.
Answer:
Page No 8.11:
Question 14:
Divide x4 + x2 + 1 by x2 + x + 1.
Answer:
Page No 8.11:
Question 15:
Divide x5 + x4 + x3 + x2 + x + 1 by x3 + 1.
Answer:
Page No 8.11:
Question 16:
Divide 14x3 − 5x2 + 9x − 1 by 2x − 1 and find the quotient and remainder
Answer:
Quotient = 7x2 + x + 5Remainder = 4
Page No 8.11:
Question 17:
Divide 6x3 − x2 − 10x − 3 by 2x − 3 and find the quotient and remainder.
Answer:
Quotient = 3x2+ 4x + 1 Remainder = 0
Page No 8.11:
Question 18:
Divide 6x3+ 11x2 − 39x − 65 by 3x2 + 13x + 13 and find the quotient and remainder.
Answer:
Quotient = 2x-5Remainder =0
Page No 8.12:
Question 19:
Divide 30x4 + 11x3 − 82x2 − 12x + 48 by 3x2 + 2x − 4 and find the quotient and remainder.
Answer:
Quotient =10x
2-3x-12Remainder= 0
Page No 8.12:
Question 20:
Divide 9x4 − 4x2 + 4 by 3x2 − 4x + 2 and find the quotient and remainder.
Answer:
∴ Quotient = 3x2 + 4x + 2 andremainder = 0 .
Page No 8.12:
Question 21:
Verify the division algorithm i.e. Dividend = Divisor × Quotient + Remainder, in each of the following. Also, write the quotient and remainder.
Dividend | Divisor | |
(i) | 14x2 + 13x − 15 | 7x − 4 |
(ii) | 15z3 − 20z2 + 13z − 12 | 3z − 6 |
(iii) | 6y5 − 28y3 + 3y2 + 30y − 9 | 2y2 − 6 |
(iv) | 34x − 22x3 − 12x4 − 10x2 − 75 | 3x + 7 |
(v) | 15y4 − 16y3 + 9y2 − 103y + 6 | 3y − 2 |
(vi) | 4y3 + 8y + 8y2 + 7 | 2y2 − y + 1 |
(vii) | 6y5 + 4y4 + 4y3 + 7y2 + 27y + 6 | 2y3 + 1 |
Answer:
(i)
Quotient = 2x + 3
Remainder = -3
Divisor = 7x- 4
Divisor × Quotient + Remainder = (7x - 4) (2x + 3) - 3
= 14x2 + 21x - 8x - 12 - 3
= 14x2 + 13x - 15
= Dividend
Thus,
Divisor × Quotient + Remainder = Dividend
Hence verified.
(ii)
Quotient = 5z2+103z+11Remainder = 54Divisor = 3z-6Divisor × Quotient +Remainder = (3z-6) 5z2+10 3z+11+54 = 15z3+10z2+33z-30z2-20z-66+54 = 15z3-20z2+13z-12 = DividendThus,Divisor × Quotient + Remainder = Dividend
Hence verified.
(iii)
Quotient = 3y3-5y+32
Remainder = 0
Divisor = 2y2 - 6
Divisor × Quotient + Remainder =
(2y2-6) 3y3-5y +32+0=6y5-10y3+3y2-18y3+30y-9=6y5-28 y3+3y2+30y-9
= Dividend
Thus, Divisor × Quotient + Remainder = Dividend
Hence verified.
(iv)
Quotient = - 4x3 + 2x2- 8x + 30
Remainder = - 285
Divisor = 3x + 7
Divisor × Quotient + Remainder = (3x + 7) (- 4x3 + 2x2 - 8x + 30) - 285
= - 12x4 + 6x3 - 24x2 + 90x - 28x3 + 14x2- 56x + 210 - 285
= - 12x 4- 22x3- 10x2 + 34x - 75
= Dividend
Thus,
Divisor × Quotient + Remainder = Dividend
Hence verified.
(v)
Quotient = 5y3-2y2+53y
Remainder = 6
Divisor = 3y - 2
Divisor × Quotient + Remainder = (3y - 2) (5y3 - 2y2 + 53y) + 6
= 15y 4-6y3+5y2-10y3+4y2-103y+6
= 15y4-16y3+9y2-103y+6
= Dividend
Thus,
Divisor × Quotient + Remainder = Dividend
Hence verified.
(vi)
Quotient = 2y + 5
Remainder = 11y + 2
Divisor = 2y2- y + 1
Divisor × Quotient + Remainder = (2y2 - y + 1)(2y + 5) + 11y + 2
= 4y3 +10y2 - 2y2 - 5y + 2y + 5 + 11y + 2
= 4y3 + 8y2 + 8y + 7
= Dividend
Thus,
Divisor × Quotient + Remainder = Dividend
Hence verified.
(vii)
Quotient = 3y2 + 2y + 2
Remainder = 4y2 + 25y + 4
Divisor = 2y3 + 1
Divisor ×
Quotient + Remainder = (2y3 + 1)(3y2 + 2y + 2) + 4y2 + 25y + 4
= 6y5
+ 4y4 + 4y3 + 3y2 + 2y + 2 + 4y2 + 25y + 4
= 6y5
+ 4y4 + 4y3 + 7y2 + 27y + 6
= Dividend
Thus,
Divisor × Quotient + Remainder = Dividend
Hence verified.
Page No 8.12:
Question 22:
Divide 15y4 + 16y3 + 103y − 9y2 − 6 by 3y − 2. Write down the coefficients of the terms in the quotient.
Answer:
∴ Quotient = 5y3 + (26/3)y2 + (25/ 9)y + (80/27)
Remainder = (- 2/27)
Coefficient of y3 = 5
Coefficient of y2 = (26/3)
Coefficient of y = (25/9)
Constant = (80/27)
Page No 8.12:
Question 23:
Using division of polynomials, state whether
(i) x + 6 is a factor of x2 − x − 42
(ii) 4x − 1 is a factor of 4x2 − 13x − 12
(iii) 2y − 5 is a factor of 4y4 − 10y3 − 10y2 + 30y − 15
(iv) 3y2 + 5 is a factor of 6y5 + 15y4 + 16y3 +
4y2 + 10y − 35
(v) z2+ 3 is a factor of z5 − 9z
(vi) 2x2 − x + 3 is a factor of 6x5 − x4 + 4x3 − 5x2 − x − 15
Answer:
(i)
Remainder is zero. Hence (x+6) is a factor of x2 -x-42
(ii)
As the remainder is non zero . Hence ( 4x-1) is not a factor of 4x2 -13x-12
(iii)
∵ The remainder is non zero,
2y - 5
is not a factor of 4y4-10y3-10y2+30y-15.
(iv)
Remainder is zero. Therefore, 3y2 + 5 is a factor of 6y5+15y4+16y3+4y 2+10y-35.
(v)
Remainder is zero; therefore, z2 + 3 is a factor of z 5 -9z.
(vi)
Remainder is zero ; therefore, 2x2 -x+3 is a factor of 6x5-x4 +4x3-5x2-x-15.
Page No 8.12:
Question 24:
Find the value of a, if x + 2 is a factor of 4x4 + 2x3 − 3x2 + 8x + 5a.
Answer:
We have to find the value of a if (x+2) is a factor of (4x4+2x3-3x2+8x+5a).Substituting x=-2 in 4x4+2x3-3x2+8x+ 5a, we get:4(-2)4+2(-2)3-3(-2)2+8(-2)+5a =0or, 64-16-12-16+5a=0or, 5a=-20or, a= -4∴ If (x+2) is a factor of (4x4+2x3-3x2+8x+5a), a=-4.
Page No 8.12:
Question 25:
What must be added to x4 + 2x3 − 2x2 + x − 1 , so that the resulting polynomial is exactly divisible by x2+ 2x − 3?
Answer:
Thus, (x - 2) should be added to (x4+2x3-2x2+x-1) to make the resulting polynomial exactly divisible by (x2+2x-3).
Page No 8.15:
Question 1:
Divide
the first polynomial by the second in each of the following. Also, write the quotient and remainder:
(i) 3x2 + 4x + 5, x − 2
(ii) 10x2 − 7x + 8, 5x − 3
(iii) 5y3 − 6y2 + 6y − 1, 5y − 1
(iv) x4 − x3 + 5x, x − 1
(v) y4 + y2, y2 −
2
Answer:
(i) 3x2+4x+5x-2=3x(x-2)+10(x-2)+25(x-2)=(x-2
)(3x+10)+25(x-2)=(3x+10)+25(x-2)
Therefore, quotient=3x+10 and remainder=25.(ii) 10
x2-7x+85x-3=2x(5x-3)-15(5
x-3)+475(5x-3)=(5x-3)(2x-1
5)+475(5x-3)=(2x-15)+4755x-3Therefore, quotient=2x-15 and remainder=475.
(iii) 5y3-6y2+6y-15y-1=
y2(5y-1)-y(5y-1)+1(5y-1)(5y-1)
=(5y-1)(y2-y+1)(5y-1)=(y2-y
+1)Therefore, Quotient = y2-y+1 and remainder = 0
(iv) x4-x3+5xx-1=x3(x-1)+5(x-1)+5x-1
=(x-1)(x3+5)+5x-1=(x3+5)+5x-1Therefore, quotient = x3
+5 and remainder = 5.
(v) y4+y2y2
-2=y2(y2-2)+3(y2-2)+6y2-2=(y2-2)(y2+3)+6y2-2=(y2+3)
+6y2-2Therefore, quotient = y2+3 and remainder = 6.
Page No 8.15:
Question 2:
Find whether the first polynomial is a factor of the second.
(i) x + 1, 2x2 + 5x + 4
(ii) y − 2, 3y3 + 5y2 + 5y + 2
(iii) 4x2 − 5, 4x4 + 7x2 + 15
(iv) 4 − z, 3z2 −
13z + 4
(v) 2a − 3, 10a2 − 9a − 5
(vi) 4y + 1, 8y2 − 2y + 1
Answer:
(i) 2x2+5x+4 x+1=2x(x+1)+3(x+1)+1x+1= (x+1)(2x+3)+1(x+1)=(2x+3)+1x+1 ∵ Remainder=1Therefore, (x+1) is not a factor of 2x2+5x+4
(ii) 3y3+5y2+5y+2y-2=3y2(y-2)+11y(y-2)+27(y-2)+56y-2=(y-2)(3y2+11y+27)+56y-2=(3y2+11y+27)+56y-2∵ Remainder = 56∴ (y-2) i s not a factor of 3y3+5y2+5y+2.
(iii) 4x4+2+154x2-5= x2(4x2-5)+3(4x2-5)+304x2-5= (4x2-5)(x2+3)+304x2-5=(x2+3)+304x2-5∵ Remainder = 30Therefore, (4x2-5) is not a factor of 4x4+7x2+15
(iv) 3z2-13z+44-z= 3z2-12z-z+44-z=3z(z-4)-1(z-4)4-z=(z-4)(3z-1)4-z=(4-z)(1-3z)4-z=1-3z∵ Remainder = 0∴ (4-z) is a factor of 3z2-13z+4.
(V) 10a2 -9a-52a-3=5a(2a-3)+3(2a-3)+42a-3=(2a-3)(5a+3)+42a-3=(5a+3)+42a-3∵ Remainder = 4 ∴ ( 2a-3) is not a factor of 10a2-9a-5.
(vi) 8y2-2y+1 4y+1=2y(4y+1)-1(4y+1)+24y+1=(4y+1)(2y-1)+24y+1=(2y-1)+24y+1∵ Remainder = 2∴ (4y+1) is not a factor of 8y2-2y+1.
Page No 8.17:
Question 1:
Divide:
x2 − 5x + 6 by x − 3
Answer:
x2-5x+6x-3=x2-3x-2x+6x-3=x(x- 3)-2(x-3)(x-3)= (x-3)(x-2)(x-3)= x-2
Page No 8.17:
Question 2:
Divide:
ax2 − ay2 by ax + ay
Answer:
ax2- ay2ax+ay=a(x2-y2)a(x+y)=a(x+y)(x-y)a(x +y)= x-y
Page No 8.17:
Question 3:
Divide:
x4 − y4 by x2 − y2
Answer:
x4-y4x2-y2=(x2)2-(y2)2(x2-y2)=(x2+y2)(x2-y2)(x2-y2)= x2+y 2
Page No 8.17:
Question 4:
Divide:
acx2 + (bc + ad)x + bd by (ax + b)
Answer:
acx2+(b c+ad)x+bd(ax+b)=acx2+b cx+adx+bd(ax+b)=cx(ax+b) +d(ax+b)(ax+b)=(ax+b)(cx+d)(ax+b)= cx+d
Page No 8.17:
Question 5:
Divide:
(a2 + 2ab + b2) − (a2 + 2ac + c2) by 2a + b + c
Answer:
(a2+2ab+b2)-(a2+2ac+c2)(2a+b+c)=(a+b)2-(a +c)2(2a+b+c)=(a+b+a+c)(a+b-a-c)(2a+b+c) =(2a+b+c)(b-c)(2a+b+c)=b-c
Page No 8.17:
Question 6:
Divide:
14x2-12x
-12 by 12x-4
Answer:
14x2-12x-1212x-4=12x(12x-4)+3(12x-4)12x-4=( 12x-4)(12x+3)(12x-4)=12x+3
Page No 8.2:
Question 1:
Write the degree of each of the following polynomials.
(i) 2x2 + 5x2 − 7
(ii) 5x2 − 3x + 2
(iii) 2x + x2 − 8
(iv) 12
y7-12y6+48y5-10
(v) 3x3 + 1
(vi) 5
(vii) 20x3 + 12x2y2 − 10y2 + 20
Answer:
(i ) Correction : It is 2x3+5x2-7 instead of 2x2+5x2-7 . The degree of the polymonial 2x3+5x2-7 is 3. (ii) The degree of the polymonial 5x2-35x+2 is 2.(iii) The degree of the polymonial 2x+x2-8 is 2.(iv) The degree of the polymonial 12y7-12y6+48y5-10 is 7.(v) The degree of the polymonial 3x3+1 is 3.(vi) 5 is a constant polynomial and its degree is 0.(vii) The degree of the polymonial 20x3+12x2 y2-10y2+20 is 4.
Page No 8.2:
Question 2:
Which of the following expressions are not polynomials?
(i) x2 + 2x−2
(ii) ax+
x2-x3
(iii) 3y3 − 5y + 9
(iv) ax1/2 + ax + 9x2 + 4
(v) 3x−2 + 2x−1 + 4x +5
Answer:
(i) x2+2x-2 is not a polynomial because -2 is the power of variable x is not a non negative integer.(ii) ax+x2-x3 is not a polynomial because 12 is the power of variable x is not a non negative integer.(iii) 3y3-5y+9 is a polynomial because the powers of variable y are non negative integers.(iv) ax12+ax+9 x2+4 is not a polynomial because 12 is the power of variable x is not a non negative integer.(v) 3x-2+2x-1+4x+5 is not a polynomial because -2 and -1 are the powers of variable x are not non negative integers.
Page No 8.2:
Question 3:
Write each of the following polynomials in the standard form. Also, write their degree.
(i) x2 + 3 + 6x + 5x4
(ii)a2 + 4 + 5a6
(iii) (x3 −
1)(x3 − 4)
(iv) (y3− 2)(y3+ 11)
(v) a3-38a3+1617
(vi) a+
34a+43
Answer:
(i) Standard form of the given polynomial can be expressed as:(5x4+x2+6x+3) or (3+6x+ x2+5x4) The degree of the polynomial is 4.(ii) Standard form of the given polynomial can be expressed as:(5a6+a2+4) or (4+a2+5a6) The degree of the polynomial is 6.(iii) (x3-1)(x3-4)=x6-5x3+4Standard form of the given polynomial can be expressed as:(x6-5x3+4) or ( 4-5x3+x6)The degree of the polynomial is 6.(iv) (y3-2)(y3+11)=y6+9y3-22Standard form of the given polynomial can be expressed as:(y6+9y3-22) or (-22+9y 3+y6)The degree of the polynomial is 6.(v) (a 3-38)(a3+1617)=a6+77136a3-617Standard form of the given polynomial can be expressed as:(a6+77136a3-617) or (-617+77136a3+a6)The degree of the polynomial is 6.(vi) (a+34)(a+43)=a2+2512a+1 Standard form of the given polynomial can be expressed as:(a2+2512a+1 ) or (1+2512a+a2)The degree of the polynomial is 2.
Page No 8.4:
Question 1:
Divide 6x3y2z2 by 3x2yz.
Answer:
6x3y2z23x2yz=6 ×x×x×x×y×y×z×z3×x×x×y×z = 2x(3-2)y(2-1)z (2-1)=2xyz
Page No 8.4:
Question 2:
Divide 15m2n3 by 5m2n2.
Answer:
15m2n3 5m2n2=15×m×m×n×n×n5×m×m×n×n=3m(2 -2)n(3-2)=3m0n1=3n
Page No 8.4:
Question 3:
Divide 24a3b3 by −8ab.
Answer:
24a3b3-8ab= 24×a×a×a×b×b×b-8×a ×b=-3a(3-1)b(3-1)=-3a2b2
Page No 8.4:
Question 4:
Divide −21abc2 by 7abc.
Answer:
-21abc27abc= -21×a×b×c× c7×a×b×c=-3a(1-1)b(1-1)c(2-1)=-3 c
Page No 8.4:
Question 5:
Divide 72xyz2 by −9xz.
Answer:
72xyz2-9xz=72×x× y×z×z-9×x×z=-8x(1-1)yz(2-1)=-8yz
Page No 8.4:
Question 6:
Divide −72a4b5c8 by −9a2b2c3.
Answer:
-72a4b5c8-9a 2b2c3=-72×a×a×a×a×b×b×b×b×b×c×c×c×c×c×c×c ×c-9×a×a×b×b×c×c×c=8a(4-2)b(5-2)c(8-3 )=8a2b3c5
Page No 8.4:
Question 7:
Simplify:
16m3y24m2y
Answer:
16m3y24m2y=16×m×m×m×y×y4×m ×m×y=4m(3-2)y(2-1)=4my
Page No 8.4:
Question 8:
Simplify:
32m2n3p24mnp
Answer:
32m2n3p24mnp =32×m×m×n×n×n×p×p4×m×n×p=8m(2-1)n( 3-1)p(2-1)=8mn2p
Page No 8.6:
Question 1:
Divide x + 2x2 + 3x4 − x5 by 2x.
Answer:
x+2x2+3x4-x52x=x2x+2x22x +3x42x-x52x=12+x+32x3-12x4
Page No 8.6:
Question 2:
Divide y4-3 y3+12y2 by 3y.
Answer:
y4-3y3 +12y23y=y43y-3 y33y+12y23y=13 y(4-1)-y(3-1)+16y(2-1)=13y3-y2 +16y
Page No 8.6:
Question 3:
Divide −4a3 + 4a2 + a by 2a.
Answer:
-4 a3+4a2+a2a=-4a32a+4a22a+a2a=-2a (3-1)+2a(2-1)+1 2=-2a2+2a+12
Page No 8.6:
Question 4:
Divide -x6+2x4+4x3+2x2 by 2x2.
Answer:
-x6+2x4+4x3+2x22
x2=-x62x2+2x4
2x2+4x32x2+2x22x2=-12x(6-2)+2x(4-2)+22x(
3-2)+2x(2-2)=-12x4+2x2+22x+2
Page No 8.6:
Question 5:
Divide 5z3 − 6z2 + 7z by 2z.
Answer:
5z3-6z2+7z2z
=5z32z-6z22z+7z2z=52z(3-1)-3
z(2-1)+72=52z2
-3z+72
Page No 8.6:
Question 6:
Divide 3 a4+23 a3+3a2-6a by 3a.
Answer:
3a4+23a3+3a2-6a3
a=3a43a+23a33a
+3a23a-6a3a=13a(4-1)+23a(3-1
)+a(2-1)-2=13a
3+23a2+a-2
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