DAV Class 8 Maths Solution Chapter 8 Polynomials
DAV School Books Class 8 Maths Solution Chapter 8 polynomials all Question Answer.
DAV Class 8 8th chapter Polynomials full Chapter explanation is provided by expert teacher. DAV class 8 maths chapter wise solution can be found here.
DAV School Books Class 8 Maths Solution Chapter 8 Polynomials:
Worksheet 1
1) Find out whether the given expression is a polynomial or not. If not, give reasons.
i) 4x3 – 4x2+ (1/2)
ii) √3z2 – 5√z+6
iii) 6x4+(2/3)x3 – (3/4)x2 – 1
iv) 7x(2/3) – 8x(3/2)+x2
v) 5x – (1/x)+(1/x2) – 2
vi) p4 – 3p3 – p+1
Ans:
i) 4x3 – 4x2+ (1/2)
The exponents of the variable x in this equation are 3 and 2
The exponents of the variable in this equation are non-negative integers.
∴ The given expression given above is a polynomial.
ii) √3z2 – 5√z+6
The exponents of the variable z in this equation are 2 and (1/2).
In the given expression the term 5√z is not a polynomial term. The exponent of the z is (1/2), which is not an integer.
For an expression to be polynomial the exponents of the variables in the expression must be positive integers.
That’s why the given expression is not a polynomial.
iii) 6x4+(2/3)x3 – (3/4)x2 – 1
The exponents of the variable x in this equation are 4,3 and 2
The exponents of the variable in this equation are non-negative integers.
∴ The given expression given above is a polynomial
iv) 7x(2/3) – 8x(3/2)+x2
The exponents of the variable x in this equation are (2/3), (3/2) and 2.
In the given expression the terms x(2/3) and x(3/2) are not polynomial terms. The exponents of the x are (2/3) and (3/2), which are not positive integers.
For an expression to be polynomial the exponents of the variables in the expression must be non-negative integers.
That’s why the given expression is not a polynomial.
v) 5x – (1/x)+(1/x2) – 2
If we simplify this expression we get 5x – x-1+x-2 – 2
The exponents of the variable x in this equation are 1, (-1) and (-2).
In the given expression the terms x(-1) and x(-2) are not polynomial terms. The exponents of the x are (-1) and (-2), which are not positive integers.
For an expression to be polynomial the exponents of the variables in the expression must be non-negative integers.
That’s why the given expression is not a polynomial.
vi) p4 – 3p3 – p+1
The exponents of the variable p in this equation are 4, 3 and 1
The exponents of the variable in this equation are non-negative integers.
∴ The given expression given above is a polynomial.
2) Write each of the following polynomials in standard form and also write down their degree.
i) p6 – 8p9+p7+5
ii) 4z3 – 3z5+2z4+z+1
iii) [x+ (2/3)] × [x+ (3/4)]
iv) [x2 – (2/3)] × [x2+(4/3)]
v) (z2+5) × (z2 – 6)
vi) (y3 – 4) × (y3 – 5)
vii) (p2+2) × (p2+7)
viii) [(5/6)z – (3/4)z2 – (2/3)z3+1]
ix) 4p+15p6 – p5+4p2+3
x) q10+q6 – q4+q8
Ans:
i) p6 – 8p9+p7+5
If we write this expression in standard form we get
–8p9+p7+p6+5
The highest power of the variable p is 9
∴ –8p9+p7+p6+5 is polynomial in p of degree 9.
ii) 4z3 – 3z5+2z4+z+1
If we write this expression in standard form we get
–3z5+2z4+4z3+1
The highest power of the variable z is 5
∴ –3z5+2z4+4z3+1 is polynomial in z of degree 5.
iii) [x+ (2/3)] × [x+ (3/4)]
If we write this expression in standard form we get
x2+ [(2/3) + (3/4)]x + (2/3)(3/4)
= x2+ (17/12) x+ (1/2)
The highest power of the variable x is 2
∴ x2+ (17/12) x+ (1/2) is polynomial in x of degree 2.
iv) [x2 – (2/3)] × [x2+(4/3)]
If we write this expression in standard form we get
x4+ [(4/3) – (2/3)]x2 – (2/3)(4/3)
= x4+ (2/3) x2+ (8/9)
The highest power of the variable x is 4
∴ x4+ (2/3) x2+ (8/9) is polynomial in x of degree 4.
v) (z2+5) × (z2 – 6)
If we write this expression in standard form we get
z4+ [5 – 6] z2 – (5×6)
= z4 – z2+ 30
The highest power of the variable z is 4
∴ z4 – z2+ 30 is polynomial in z of degree 4.
vi) (y3 – 4) × (y3 – 5)
If we write this expression in standard form we get
y6 – [4+5] y3 + (4×5)
= y6 – 9y3+ 20
The highest power of the variable y is 6
∴ y6 – 9y3+ 20 is polynomial in y of degree 6.
vii) (p2+2) × (p2+7)
If we write this expression in standard form we get
p4+ [2 + 7] p2 + (2×7)
= p4 + 9p2+ 14
The highest power of the variable p is 4
∴ p4 + 9p2+ 14 is polynomial in p of degree 4.
viii) [(5/6)z – (3/4)z2 – (2/3)z3+1]
If we write this expression in standard form we get
[– (2/3)z3 – (3/4)z2 + (5/6)z +1]
The highest power of the variable z is 3
∴ [– (2/3)z3 – (3/4)z2 + (5/6)z +1] is polynomial in z of degree 3.
ix) 4p+15p6 – p5+4p2+3
If we write this expression in standard form we get
15p6– p5+4p2+4p+3
The highest power of the variable p is 6
∴ 15p6– p5+4p2+4p+3 is polynomial in p of degree 6.
x) q10 +q6 – q4+q8
If we write this expression in standard form we get
q10 +q8+q6 – q4
The highest power of the variable q is 10
∴ q10 +q8+q6 – q4 is polynomial in q of degree 10.
3) Divide the following monomials by the given monomials:
i) 6x3 by 3x2
ii) – 35x4 by – 7x3
iii) – 5z2 by √5z
iv) 16p4 by – 6p2
v) 4√2y3 by 3√2y2
vi) (3/4)p2 by (4/3)p2
Ans:
i) 6x3 by 3x2
Now, 6x3/3x2 = (6/3) × (x3/x2) = 2x(3-2) = 2x
ii) – 35x4 by – 7x3
Now, (– 35x4) / (– 7x3) = (35/7) × (x4/x3) = 5x(4-3) = 5x
iii) – 5z2 by √5z
Now, (– 5z2)/(√5z) = (–5/√5) × (z2/z) = –√5z(2-1) = –√5z
iv) 16p4 by – 6p2
Now, (16p4)/(– 6p2) = (–16/6) × (p4/p2) = – (8/3)p2
v) 4√2y3 by 3√2y2
Now, (4√2y3)/(3√2y2) = (4√2/3√2) × (y3/y2) = (4/3)y(3-2) = (4/3)y
vi) (3/4)p2 by (4/3)p2
Now, [(3/4)p2]/[(4/3)p2] = [(3/4) ÷ (4/3)] × [p2/p2] = (9/16) p(1-1) = 9/16
4) Divide the following polynomial by a monomial:
i) 6x4 – 24x3+15x2+9 by (-3x2)
ii) -12x+22x2 – 16x3+4 by 2x
iii) – 8p3+16p2+14p+ 1 by 4p
iv) – 3x2+√3x by √3x
v) (2/3)z4– (1/3)z2–1 by (1/3)z
vi) 2√2q4+4√2q3+q2 by (–2√2q2)
Ans:
i) 6x4 – 24x3+15x2+9 by (-3x2)
Now, (6x4 – 24x3+15x2+9) / (-3x2)
= (6x4/-3x2) – (24x3/-3x2) + (15x2/-3x2) + (9/-3x2)
= (-6/3)×(x4/x2) – (–24/3)×(x3/x2) + (–15/3)(x2/x2) + (–9/3)(1/x2)
= –2x2+8x–5–3x-2
ii) -12x+22x2 – 16x3+4 by 2x
Now, (–12x+22x2 – 16x3+4) / 2x
= (–12x/2x) + (22x2/2x) – (16x3/2x) + (4/2x)
= (–12/2) × (x/x) + (22/2) × (x2/x) – (16/2) × (x3/x) + 2x-1
= –6x0+ 11x–8x2+2x-1
=–8x2+ 11x –6 +2x-1
iii) – 8p3+16p2+14p+ 1 by 4p
Now, (– 8p3+16p2+14p+ 1) / 4p
= (– 8p3/4p) + (16p2/4p) + (14p/4p) + 1/4p
= (–8/4) × (p3/p) + (16/4) × (p2/p) + (14/4) × (p/p) + p-1/4
= –2p2+4p+3.5+0.25p-1
iv) – 3x2+√3x by √3x
Now, (– 3x2+√3x)/√3x
= (– 3x2/√3x) + (√3x/√3x)
= (–3/√3)×(x2/x) + (√3x)(√3x)
= –√3x + 1
v) (2/3)z4– (1/3)z2–1 by (1/3)z
Now, [(2/3)z4– (1/3)z2–1]/[(1/3)z]
= [(2/3)z4] ÷ [(1/3)z] – [(1/3)z2] ÷ [(1/3)z] – [1÷ [(1/3)z]]
= [(2/3) ÷ (1/3)] × (z4/z) – [(1/3) ÷ (1/3)] × (z2/z) – 3z-1
= 2z3 – z – 3z-1
vi) 2√2q4+4√2q3+q2 by (–2√2q2)
Now, (2√2q4+4√2q3+q2) / (–2√2q2)
= (2√2q4/–2√2q2) + (4√2q3/–2√2q2) + (q2/–2√2q2)
= (–2√2/2√2) × (q4/q2) + (–4√2/2√2) × (q3/q2) + (–1/2√2) × (q2/q2)
= – q(4-2) – 2q – (1/2√2)
= –q2 – 2q – (1/2√2)
Worksheet 2
1) Using factor method, divide the following polynomials by a binomial.
i) x2+3x+2 by (x+1)
ii) x2–7x–18 by (x–9)
iii) p2+6p+8 by (p+2)
iv) p2–p–42 by (p+6)
v) y2+12y+35 by (y+7)
vi) z2–10z+16 by (z–2)
vii) x4+3x2–10 by (x2+5) (Hint: x2 = y)
viii) x2–7x+12 by (x–4)
Ans:
i) x2+3x+2 by (x+1)
To solve this problem by using factor method first we have to factorize x2+3x+2
x2+3x+2 = x2+2x+x+2 = x(x+2)+1(x+2) = (x+2)(x+1)
[∵ x2+ (a+b) x + ab = (x+a)(x+b)]
Now, (x2+3x+2)/(x+1)
= [(x+2)(x+1)] / (x+1)
= (x+2)
ii) x2–7x–18 by (x–9)
To solve this problem by using factor method first we have to factorize x2–7x–18
x2–7x–18 = x2–9x + 2x–18 = x(x-9) +2(x – 9) = (x–9)(x+2)
Now, (x2–7x–18)/(x–9)
= [(x–9)(x+2)] / (x – 9)
= (x+2)
iii) p2+6p+8 by (p+2)
To solve this problem by using factor method first we have to factorize p2+6p+8
p2+6p+8 = p2+4p+2p+8 = p(p+4)+2(p+4)=(p+4)(p+2)
Now, (p2+6p+8)/(p+2)
= [(p+4)(p+2)]/(p+2)
= (p+4)
iv) p2–p–42 by (p+6)
To solve this problem by using factor method first we have to factorize p2– p – 42
p2–p–42 = p2 – 7p+6p – 42 = p(p – 7)+6 (p – 7) = (p–7)(p+6)
Now, (p2–p–42)/(p+6)
= [(p–7)(p+6)]/(p+6)
= (p–7)
v) y2+12y+35 by (y+7)
To solve this problem by using factor method first we have to factorize y2+12y+35
y2+12y+35 = y2+7y+5y+35 = y(y+7)+5(y+7) = (y+7)(y+5)
Now, (y2+12y+35) / (y+7)
= (y+7)(y+5)/(y+7)
= (y+5)
vi) z2–10z+16 by (z–2)
To solve this problem by using factor method first we have to factorize z2–10z+16
z2–10z+16 = z2 – 8z – 2z + 16 = z(z – 8) – 2(z – 8) = (z – 8)(z – 2)
Now, (z2–10z+16)/(z – 2)
= (z–8)(z–2)/(z–2)
= (z–8)
vii) x4+3x2–10 by (x2+5) (Hint: x2 = y)
Let us consider x2=y
(x4+3x2–10)/(x2+5)
(y2+3y–10)/(y+5)
To solve this problem by using factor method first we have to factorize y2+3y–10
y2+3y–10 = y2+5y–2y–10 = y(y+5) – 2(y+5) = (y+5)(y–2)
Now, (y2+3y–10)/(y+5)
= (y+5)(y–2)/(y+5)
= (y–2)
= x2 – 2
viii) x2–7x+12 by (x–4)
To solve this problem by using factor method we have to factorize x2–7x+12
x2–7x+12 = x2–4x–3x+12 = x(x – 4) – 3(x – 4) = (x – 4)(x – 3)
Now, (x2–7x+12) / (x – 4)
= (x – 4)(x – 3)/(x – 4)
= (x – 3)
2) Using long division method, divide the following polynomials by a binomial and Check your answer.
i) 4p3–4p2+6p–(5/2) by (2p–1)
ii) x – 6 + 15x2 by (2+3x)
iii) 4x3 – 37x2+52x – 15 by (4x – 5)
iv) x3 – 3x2+3x – 1 by (x – 1)
v) 29z – 6z2 – 28 by (3z – 4)
vi) 12x + 10x3+8x4+15 by (5+4x)
vii) 12z3+4z+3z2+1 by (4z+1)
viii) 8x2 – 2 – 3x + 12x3 by (4x2 – 1)
Ans:
i) 4p3–4p2+6p–(5/2) by (2p–1)
We know that when the remainder is zero then the Dividend = Divisor × Quotient
So to check if the answer we got from long division method we will use this relation.
∴ 4p3–4p2+6p–(5/2) = (2p-1) × [2p2–p+ (5/2)]
Now L.H.S = 4p3–4p2+6p–(5/2)
R.H.S = (2p-1) × [2p2–p+ (5/2)]
= 4p3–2p2+5p – 2p2+p – (5/2)
= 4p3– 4p2+ 6p – (5/2)
∴ L.H.S = R.H.S
ii) x – 6 + 15x2 by (2+3x)
Before solving this problem we have to write dividend and the divisor in the decreasing order of the powers of the variables.
Dividend = 15x2+x– 6
Divisor = 3x + 2
We know that when the remainder is zero then the Dividend = Divisor × Quotient
So to check if the answer we got from long division method we will use this relation.
∴15x2+x– 6 = (3x + 2) × [5x – 3]
Now L.H.S = 15x2+x– 6
R.H.S = (3x + 2) × [5x – 3]
= 15x2– 9x+10x – 6
= 15x2 + x – 6
∴L.H.S = R.H.S
iii) 4x3 – 37x2+52x – 15 by (4x – 5)
We know that when the remainder is zero then the Dividend = Divisor × Quotient
So to check if the answer we got from long division method we will use this relation.
∴4x3 – 37x2+52x – 15 = (4x – 5) × (x2 – 8x+3)
Now L.H.S = 4x3 – 37x2+52x – 15
R.H.S = (4x – 5) × (x2 – 8x+3)
= 4x3 – 32x2+12x – 5x2+40x – 15
= 4x3 – 37x2+52x – 15
∴L.H.S = R.H.S
iv) x3 – 3x2+3x – 1 by (x – 1)
We know that when the remainder is zero then the Dividend = Divisor × Quotient
So to check if the answer we got from long division method we will use this relation.
∴x3 – 3x2+3x – 1 = (x – 1) × (x2– 2x + 1)
Now L.H.S = x3 – 3x2+3x – 1
R.H.S = (x – 1) × (x2– 2x + 1)
= x3 – 2x2+x – x2+2x – 1
= x3 – 3x2+3x – 1
∴L.H.S = R.H.S
v) 29z – 6z2 – 28 by (3z – 4)
Before solving this problem we have to write dividend and the divisor in the decreasing order of the powers of the variables.
Dividend = – 6z2 + 29z– 28
Divisor = 3z – 4
We know that when the remainder is zero then the Dividend = Divisor × Quotient
So to check if the answer we got from long division method we will use this relation.
∴– 6z2 + 29z– 28 = (3z – 4) × (– 2z + 7)
Now L.H.S = – 6z2 + 29z– 28
R.H.S = (3z – 4) × (- 2z + 7)
= – 6z2+ 21z+ 8z – 28
= – 6z2+29z –28
∴L.H.S = R.H.S
vi) 12x + 10x3+8x4+15 by (5+4x)
Before solving this problem we have to write dividend and the divisor in the decreasing order of the powers of the variables.
Dividend = 8x4+10x3+12x+15
Divisor = 4x+5
We know that when the remainder is zero then the Dividend = Divisor × Quotient
So to check if the answer we got from long division method we will use this relation.
∴8x4+10x3+12x+15 = (4x+5) × (2x3+3)
Now L.H.S = 8x4+10x3+12x+15
R.H.S = (4x+5) × (2x3+3)
= 8x4+12x + 10x3+15
= 8x4+ 10x3+12x +15
∴L.H.S = R.H.S
vii) 12z3+4z+3z2+1 by (4z+1)
Before solving this problem we have to write dividend and the divisor in the decreasing order of the powers of the variables.
Dividend = 12z3+3z2 + 4z+1
Divisor = 4z+1
We know that when the remainder is zero then the Dividend = Divisor × Quotient
So to check if the answer we got from long division method we will use this relation.
∴12z3+3z2+4z+1 = (4z+1) × (3z2+1)
Now L.H.S = 12z3+3z2+4z+1
R.H.S = (4z+1) × (3z2+1)
= 12z3+4z+3z2+1
= 12z3+3z2+4z+1
∴L.H.S = R.H.S
viii) 8x2 – 2 – 3x + 12x3 by (4x2 – 1)
Before solving this problem we have to write dividend and the divisor in the decreasing order of the powers of the variables.
Dividend = 12x3+8x2 – 3x – 2
Divisor = 4x2 – 1
We know that when the remainder is zero then the Dividend = Divisor × Quotient
So to check if the answer we got from long division method we will use this relation.
∴12x3+8x2 – 3x – 2 = (4x2 – 1) × (3x+2)
Now L.H.S = 12x3+8x2 – 3x – 2
R.H.S = (4x2 – 1) × (3x+2)
= 12x3+ 8x2–3x–2
∴L.H.S = R.H.S
3) Using long division method, show that the second polynomial is the factor of the first polynomial.
i) x4 – 16; x+2
ii) z4 – z3+3z2 – 2z + 2; z2+2
iii) 2y3 – 14y+12; y+3
iv) 4p3 – 12p2 – 37p – 15; 2p+1
v) 4q3 – 6q2 – 4q+3; 2q – 1
vi) 20x2 – 32x – 16; 5x+2
vii) 3x2 – 11x+6; x – 3
viii) 6x3 + 19x2+13x+ (– 3); 2x+3
Ans:
i) x4 – 16; x+2
Divide (x4 – 16) by (x+2)
∵ Remainder is zero .
∴(x+2) is a factor of (x4 – 16)
ii) z4 – z3+3z2 – 2z + 2; z2+2
Divide (z4 – z3+3z2 – 2z + 2) by (z2+2)
∵ Remainder is zero.
∴(z2+2) is a factor of (z4 – z3+3z2 – 2z + 2)
iii) 2y3 – 14y+12; y+3
Divide (2y3 – 14y+12) by (y+3)
∵ Remainder is zero.
∴(y+3) is a factor of (2y3 – 14y+12)
iv) 4p3 – 12p2 – 37p – 15; 2p+1
Divide (4p3 – 12p2 – 37p – 15) by (2p+1)
∵ Remainder is zero.
∴(2p+1) is a factor of (4p3 – 12p2 – 37p – 15)
v) 4q3 – 6q2 – 4q+3; 2q – 1
Divide (4q3 – 6q2 – 4q+3) by (2q – 1)
∵ Remainder is zero.
∴(2q – 1) is a factor of (4q3 – 6q2 – 4q+3)
vi) 20x2 – 32x – 16; 5x+2
Divide (20x2 – 32x – 16) by (5x+2)
∵ Remainder is zero.
∴(5x+2) is a factor of (20x2 – 32x – 16)
vii) 3x2 – 11x+6; x – 3
Divide (3x2 – 11x+6) by (x – 3)
∵ Remainder is zero.
∴(x – 3) is a factor of (3x2 – 11x+6)
viii) 6x3 + 19x2+13x+ (– 3); 2x+3
Divide [6x3 + 19x2+13x+ (– 3)] by (2x+3)
∵ Remainder is zero.
∴(2x+3) is a factor of [6x3 + 19x2+13x+ (– 3)]
Worksheet 3
1) Find the quotient and the remainder when the first polynomial is divided by the second.
i) –6a2+29a–30 by 3a–4
ii) 4p3+7 by –p+3
iii) y4–y2+4 by y2–4
iv) –6x4+5x2+11x+1 by 2x2+1
Ans:
i) –6a2+29a–30 by 3a–4
Dividend = –6a2+29a–30
Divisor = 3a–4
Quotient = (–2a+7)
Remainder = (–2)
ii) 4p3+7 by –p+3
Dividend = 4p3+7
Divisor = –p+3
Quotient = (–4p2–12p–36)
Remainder = (115)
iii) y4–y2+4 by y2–4
Dividend = y4–y2+4
Divisor = y2–4
Quotient = (y2+3)
Remainder = (16)
iv) –6x4+5x2+11x+1 by 2x2+1
Dividend = –6x4+5x2+11x+1
Divisor = 2x2+1
Quotient = (–3x2+4)
Remainder = (11x – 3)
2) Divide and verify the result by using: Dividend = Divisor × Quotient + Remainder.
i) y3–8 by y–2
ii) p4+p3–p2+1 by p–1
iii) x2–3x+7 by x–2
iv) 2x2+5x+4 by x+1
Ans:
i) y3–8 by y–2
Dividend = y3 – 8
Divisor = y – 2
We have to prove that Dividend = Divisor × Quotient + Remainder
Now,
Dividend = y3 – 8
Divisor = y – 2
Quotient = (y2+2y+4)
Remainder = 0
∴ R.H.S = Divisor × Quotient + Remainder
= (y – 2) × (y2+2y+4) + 0
= y3+2y2+4y – 2y2 – 4y – 8
= y3 – 8
= L.H.S
Hence, L.H.S = R.H.S
ii) p4+p3–p2+1 by p–1
Dividend = p4+p3–p2+1
Divisor = p–1
We have to prove that Dividend = Divisor × Quotient + Remainder
Now,
Dividend = p4+p3–p2+1
Divisor = p–1
Quotient = (p3+2p2+p+1)
Remainder = 2
∴ R.H.S = Divisor × Quotient + Remainder
= (p–1) × (p3+2p2+p+1) + 2
= p4+2p3+p2+p–p3–2p2–p–1+2
= p4+p3–p2+1
= L.H.S
Hence, L.H.S = R.H.S
iii) x2–3x+7 by x–2
Dividend = x2–3x+7
Divisor = x–2
We have to prove that Dividend = Divisor × Quotient + Remainder
Now,
Dividend = x2–3x+7
Divisor = x–2
Quotient = (x – 1)
Remainder = 5
∴ R.H.S = Divisor × Quotient + Remainder
= (x–2) × (x – 1) + 5
= x2 – x – 2x+2+5
= x2–3x+7
= L.H.S
Hence, L.H.S = R.H.S
iv) 2x2+5x+4 by x+1
Dividend = 2x2+5x+4
Divisor = x+1
We have to prove that Dividend = Divisor × Quotient + Remainder
Now,
Dividend = 2x2+5x+4
Divisor = x+1
Quotient = (2x+3)
Remainder = 1
∴ R.H.S = Divisor × Quotient + Remainder
= (x+1) × (2x+3) + 1
= 2x2+3x+2x+3+1
= 2x2+5x+4
= L.H.S
Hence, L.H.S = R.H.S
3) Find out whether or not the first polynomial is a factor of the second polynomial.
i) 4a–1, 12a2–7a–2
ii) 3y+1, 3y3+7y2+2y
iii) x–3, x3+4x2–3x+5
iv) x2+3, 4x4+7x2–15
v) p2+9, p4+13p2+36
vi) x+11, x2+9x–22
Ans:
i) 4a–1, 12a2–7a–2
To find out if the first polynomial is a factor of the second polynomial we are going to divide the second polynomial with the first polynomial. If we get zero as a remainder then we can say that the first polynomial is a factor of the second polynomial. On the other hand if we get a non-zero remainder then we can say that the first polynomial is not a factor of the second polynomial.
Dividend = 12a2 – 7a – 2
Divisor = 4a – 1
We can see that,
Quotient = 3a – 1
Remainder = (–3)
∵ Remainder is non-zero.
∴ (4a – 1) is not a factor of (12a2 – 7a – 2)
ii) 3y+1, 3y3+7y2+2y
To find out if the first polynomial is a factor of the second polynomial we are going to divide the second polynomial with the first polynomial. If we get zero as a remainder then we can say that the first polynomial is a factor of the second polynomial. On the other hand if we get a non-zero remainder then we can say that the first polynomial is not a factor of the second polynomial.
Dividend = 3y3+7y2+2y
Divisor = 3y+1
We can see that,
Quotient = y2+2y
Remainder = 0
∵ Remainder is zero.
∴ (3y+1) is a factor of (3y3+7y2+2y)
iii) x–3, x3+4x2–3x+5
To find out if the first polynomial is a factor of the second polynomial we are going to divide the second polynomial with the first polynomial. If we get zero as a remainder then we can say that the first polynomial is a factor of the second polynomial. On the other hand if we get a non-zero remainder then we can say that the first polynomial is not a factor of the second polynomial.
Dividend = x3+4x2–3x+5
Divisor = x–3
We can see that,
Quotient = x2+7x+18
Remainder = 59
∵ Remainder is non-zero.
∴ (x–3) is not a factor of (x3+4x2–3x+5)
iv) x2+3, 4x4+7x2–15
To find out if the first polynomial is a factor of the second polynomial we are going to divide the second polynomial with the first polynomial. If we get zero as a remainder then we can say that the first polynomial is a factor of the second polynomial. On the other hand if we get a non-zero remainder then we can say that the first polynomial is not a factor of the second polynomial.
Dividend = 4x4+7x2–15
Divisor = x2+3
We can see that,
Quotient = 4x2 – 5
Remainder = 0
∵ Remainder is zero.
∴ (x2+3) is a factor of (4x4+7x2–15).
v) p2+9, p4+13p2+36
To find out if the first polynomial is a factor of the second polynomial we are going to divide the second polynomial with the first polynomial. If we get zero as a remainder then we can say that the first polynomial is a factor of the second polynomial. On the other hand if we get a non-zero remainder then we can say that the first polynomial is not a factor of the second polynomial.
Dividend = p4+13p2+36
Divisor = p2+9
We can see that,
Quotient = p2+4
Remainder = 0
∵ Remainder is zero.
∴ (p2+9) is a factor of (p4+13p2+36).
vi) x+11, x2+9x–22
To find out if the first polynomial is a factor of the second polynomial we are going to divide the second polynomial with the first polynomial. If we get zero as a remainder then we can say that the first polynomial is a factor of the second polynomial. On the other hand if we get a non-zero remainder then we can say that the first polynomial is not a factor of the second polynomial.
Dividend = x2+9x–22
Divisor = x+11
We can see that,
Quotient = 3a – 1
Remainder = (–3)
∵ Remainder is zero.
∴ (x+11) is a factor of (x2+9x–22).
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