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One of the most important chapters for Class 10th CBSE Maths exam is Polynomials. We have included all the important questions, along with detailed explanations, in order to assist students in achieving higher marks in Board Exams. These can help you comprehend the kind of polynomial problems that are posed and also help you evaluate your preparation.
The video below explains this:
Polynomials Detailed Video Explanation:
Read More: Degree of polynomial
Quick Revision:
- Polynomial is made up of two words ‘Poly’ and ‘Nomial’ meaning Many and Term respectively.
- A polynomial is an expression that has one or more terms with non-zero coefficients. Any number of terms can be included in a polynomial. The degree of a polynomial is the largest power of the variable in the polynomial.
- A quadratic polynomial in x with real coefficients is of the form ax² + bx + c, where a, b, c are real numbers with a ≠ 0.
- Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomial respectively.
- The zeroes of the polynomial are the values of variables for which the polynomial becomes zero. Zeroes follow the rules of algebraic identities.
- In general, if n is a zero of p(x) = ax + b, then p(n) = an + b = 0, i.e., n = -ba . Hence, the zero of the linear polynomial ax + b is -ba = - constant term coefficient of x
- If α and β are the zeroes of the quadratic polynomial ax² + bx + c, then
- sum of zeros, α+β = -ba = - coefficient of x, coefficient of x2
- product of zeros, αβ = ca = constant term coefficient of x2
- If α, β, γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d = 0, then
- α+β+γ = – ba= - coefficient of x2, coefficient of x3
- αβ+βγ+γα = ca = coefficient of x, coefficient of x3
- αβγ = – da = - constant term coefficient of x3
- Division Algorithm for Polynomials: If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then p(x) = g(x) × q(x) + r(x)
Dividend = Divisor x Quotient + Remainder
As per the NCERT book, important problems from Chapter Polynomials for Class 10 Maths are presented below.
Read More: Pair of Linear Equations in Two Variables Important Questions
Important Polynomial Questions
Question: Find the zeroes of the quadratic polynomial 3x2 – 2 and verify the relationship between the zeroes and the coefficients.
Sol. Given quadratic polynomial is 3x2-2 Consider p(x)=3x2-2
For zeroes of polynomial p(x), put p(x)=0
⇒3x2-2=0
⇒3x2=2
⇒x2=23
⇒x=±23
Hence zeroes of polynomial p(x) are +23 and -23
Here α=23 and β=-23
Hence sum of zeroes =α+β=23-23=0
Product of zeroes =αβ=23×-23=-23
Also from the polynomial p(x)=3x2-2 sum of zeroes =-ba=03=0
Product of zeroes =ca=-23
This verifies the relation.
Question: Find all the zeros of the polynomial f(x)=2x4-3x3-3x2+6x-2, if being given that two of its zeros are 2 and -2.
Sol. 2 and -2 are the zeros. ∴(x-2)(x+2) is the factor of the given polynomial.
q(x)=2x2-3x+1
=2x2-2x-x+1
=2x(x-1)-1(x-1)
=(2x-1)(x-1)
∴ other two zeros are x=1 and x=12
Question: Quadratic polynomial 2x2-3x+1 has zeroes as α and β. Now form a quadratic polynomial whose zeroes are 3α and 3β.
Sol:α and β are the zeroes of the polynomial 2x2-3x+1
⇒α+β=-ba=-(-3)2=32
αβ=ca=12
Now, zeroes of the required polynomial are 3α and 3β
S= 3α+3β = 3(α+β) = 332 = 92P = (3α)(3β) = 9(αβ)=9×12=92
Now, required polynomial is x2-Sx+p =x2-92x+92=k22x2-9x+9, where k be any constant.
Read More: Zeros of Polynomial
Question: If one zero of the quadratic polynomial f(x)=4x2-8kx+8x-9 is negative of the other, then find the zeroes of kx2+3kx+2
Sol: f(x)=4x2-8kx+8x-9=4x2-(8k-8)x-9
Let one zero of f(x) be α then other zero be -α. So, sum of zeroes =0
⇒8k-84=0⇒8k-8=0⇒k=1
Now, other given polynomial is p(x)=kx2+3kx+2 =x2+3x+2
=(x+2)(x+1)
So, zeroes of p(x) are -1 and -2.
x2+3x+2 =x2+2x+x+2 =x(x+2)+1(x+2) =(x+2)(x+1)(x+2)(x+1) =0
or
∴ other zeroes are x=-2 and x=-1
Question: If the product of zeroes of the polynomial ax2 – 6x – 6 is 4, find the value of a. Find the sum of zeroes of the polynomial.
Sol: Let α,β be the zeroes of given polynomial p(x)=ax2-6x-6 Then,
Thus,
αβ =-6a⇒4=-6a⇒a=-64=-32a =-32
Now, sum of zeroes =6a=6-32=2×6-3=-4
Question: Find the value of ‘k’ for which the polynomial x4 + 10x3 + 25x2 + 15x + k is exactly divisible by (x + 7).
Sol: p(x)=x4+10x3+25x2+15x+k
∴(x+7) is the factor.
∴p(-7)=0
or (-7)4+10(-7)3+25(-7)2+15(-7)+k=0
2401-3430+1225-105+k=0
k=91
Question: If and are the zeros of the polynomial f (x) = x2 + px + q, find polynomial whose zeros are (+)2 and (–)2.
Sol: If(x)=x2+px+q, if α and β are zeros ∴α+β=-p and αβ=q
If zeros are (α+β)2 and (α-β)2 (α-β)2=(α+β)2-4αβ
=(-p)2-4q
(α-β)2=-p2-4q
Now sum of zeros
(α+β)2+(α-β)2=(-p)2+p2-4q
=2p2-4q
Product of zeros (α+β)2(α-β)2=(-p)2+p2-4q
=4p4-4p2q
required polynomial is x2-( sum of zeros )x+ product of zeros =x2-2p2-4qx+4p4-4p2q
=x2-2p2x-4qx+p4-4p2q
Read More: Roots of Polynomials
Question: Form a quadratic polynomial whose zeroes are 3 + √2 and 3 – √2.
Ans: Sum of zeroes,
S = (3 + √2) + (3 – √2) = 6
Product of zeroes,
P = (3 + √2) x (3 – √2) = (3)2 – (√2)2 = 9 – 2 = 7
Quadratic polynomial = x2 – Sx + P = x2 – 6x + 7
Question: Find other zeroes of the polynomial x4-7x2+12 if it is given that two of its zeroes are √3 and -√3
Ans: p(x)=x4-7x2+12
+3 and -3 are the zeroes of p(x) ∴ (x-3)(x+3) is a factor of p(x)
⇒x2-3 is a factor of p(x). For other factors of p(x), ∴x4-7x2+12=x2-3x2-4=x2-3(x-2)(x+2)
∴ Other two zeroes are 2 and -2.
Question: If α,β are the zeroes of the polynomial 2y2+7y+5, write the value of α+β+αβ.
Sol:P(y)=2y2+7y+5
α,β are zeroes of P(y)
∴ α+β =-72αβ =52α+β+αβ =-72+52=-22=-1
Question: Find a quadratic polynomial whose zeroes are 3+55 and 3-55.
Sol:Sum of zeroes (α+β),
S =3+55+3-55 =3+5+3-55=65 …α=3+55β=3-55
Product of zeroes (α×β),
P=3+553-55=9-525=425
Quadratic polynomial is x2-Sx+P=0
=x2-65x+425=25x2-30x+425 =12525x2-30x+4
or k25x2-30x+4
...where [k∈R]
Question: Find the zeroes of the quadratic polynomial 3x2-75 and verify the relationship between the zeroes and the coefficients.
Sol: We have, 3x2-75
=3x2-25 =3x2-52
=3(x-5)(x+5)
Zeroes are:
x-5=0 or x+5=0
x=5 or x=-5
Verification:
Here a=3,b=0,c=-75
Sum of the zeroes =5+(-5)=0
= – 03 = – ( Coefficient of x) Coefficient of x2= – ba
Product of the zeroes =5(-5)=-25
= – 753= Constant term Coefficient of x2=ca
Question: Find the zeroes of p(x)=2x2-x-6 and verify the relationship of zeroes with these co-efficients.
Sol: p(x)=2x2-x-6…[ Given ]
=2x2-4x+3x-6
=2x(x-2)+3(x-2)
=(x-2)(2x+3)
Zeroes are:
x-2=0 or 2x+3=0
x=2 or x=-32
Verification:
Here a=2,b=-1,c=-6
Sum of zeroes =2 +-32=4-32=12 =12=- Coefficient of x Coefficient of x2=-ba Product of zeroes =2×-32=-62 =-62= Constant term Coefficient of x2=ca
∴ Relationship holds.
Question: Find a quadratic polynomial, the sum and product of whose zeroes are 0 and -35 respectively. Hence find the zeroes.
Sol: Quadratic polynomial =x2-( Sum )x+ Product
=x2-(0)x+-35=x2-35 =(x)2-352=x-35x+35
Zeroes are, x-35=0 or
x+35=0
⇒x=35 or
x=-35
⇒x= 35× 55 or
x= -35 × 55
⇒x=155 or x=-155
The zeroes are 155 and -155.
Question: Find the zeroes of the quadratic polynomial f(x)=x2-3x-28 and verify the relationship between the zeroes and the co-efficients of the polynomial.
Sol: p(x)=x2-3x-28
=x2-7x+4x-28
=x(x-7)+4(x-7)
=(x-7)(x+4)
Zeroes are:
x-7=0 or x+4=0
x=7 or x=-4
Verification:
a=1, b=-3, c=-28 Sum of zeroes =7 +(-4)=3 =31= Coefficient of x Coefficient of x2=-ba Product of zeroes =7(-4)=-28 =-281= Constant term Coefficient of x2=ca
Relationship holds.
Question: If p(x)=x3-2x2+kx+5 is divided by (x-2), the remainder is 11 . Find k. Hence find all the zeroes of x3+kx2+3x+1.
Sol: p(x)=x3-2x2+kx+5
When x-2 p(2)=(2)3-2(2)2+k(2)+5
⇒11=8-8+2k+5
⇒11-5=2k
⇒6=2k
⇒k=3
Let q(x)=x3+kx2+3x+1
=x3+3x2+3x+1
=x3+1+3x2+3x
=(x)3+(1)3+3x(x+1)
=(x+1)3
=(x+1)(x+1)(x+1)…?a3+b3+3ab(a+b)=(a+b)3
All zeroes are:
x+1=0⇒x=-1
x+1=0⇒x=-1
x+1=0⇒x=-1
Hence zeroes are -1,-1 and -1.
Question: If a and β are zeroes of p(x)=kx2+4x+4, such that a2+β2=24, find k.
Sol: We have, p(x)=kx2+4x+4
Here a=k,b=4,c=4
Sum of zeroes, α+β=-ba=-4k
Product of zeroes, αβ=ca=4k
α2+β2=24 … [Given (α+β)2-2αβ =24-4k2-24k=24⇒16k2-8k=24 ⇒ 16-8kk2=241⇒24k2=16-8k⇒24k2+8k-16=0
⇒3k2+k-2=0….[ Dividing both sides by 8]
⇒3k2+3k-2k-2=0
⇒3k(k+1)-2(k+1)=0
⇒(k+1)(3k-2)=0
⇒k+1=0 or 3k-2=0
⇒k=-1 or k=23
Question: If a and β are the zeroes of the polynomial p(x)=2x2+5x+k, satisfying the relation, a2+β2+aβ=214 then find the value of k.
Sol: Given polynomial is p(x)=2x2+5x+k
Here a=2,b=5,c=k
α+β=-ba=-52
αβ=ca=k2
α2+β2+αβ=214 ...[Given]
(α+β)2-2αβ+αβ=214
(α+β)2-αβ=214⇒-522-k2=214
-k2=214-254=-44 ⇒ -k2=-1∴k=2
Read More: Polynomials Formulas
Question: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.
(i) x2-2x-8
(ii) 4s2-4s+1
(iii) 6x2-3-7x
(iv) 4u2+8u
(v) t2-15
(vi) 3x2-x-4
Sol:
(i) x2-2x-8
Comparing given polynomial with general form ax2+bx+c We get a=1,b=-2 and c=-8 We have, x2-2x-8 =x2-4x+2X-8
=x-4+2(x-4)=x-4x+2
Equating this equal to 0 will find values of 2 zeroes of this polynomial. x-4x+2=0
⇒x=4,-2 are two zeroes.
Sum of zeroes =4-2=2=-(-2)1=-ba=- Coefficient of x Coefficient of x2
Product of zeroes =4x-2=-8=-81=ca= Constant term Coefficient of x2
(ii) 4s2-4s+1
Here, a=4, b=-4 and c=1
We have, 4s2-4s+1
=4s2-2s-2s+1=2s(2s-1)-1(2s-1)
=2s-12s-1
Equating this equal to 0 will find values of 2 zeroes of this polynomial.
⇒(2s-1)(2s-1)=0
⇒S=12,12
Therefore, two zeroes of this polynomial are 12,12 Sum of zeroes =12+12=1=-(-1)1×44=-(-4)4=-ba=- Coefficient of x Coefficient of x2
Product of Zeroes =12×12=14=ca= Constant term Coefficient of x2
(iii) 6x2-3-7x
Here, a=6,b=-7 and c=-3 We have, 6x2-3-7x
=6x2-7x-3=6x2-9X+2X-3
=3x(2x-3)+1(2x-3)=(2x-3)3X+1
Equating this equal to 0 will find values of 2 zeroes of this polynomial. ⇒(2x-3)(3x+1)=0
⇒X=32,-13
Therefore, two zeroes of this polynomial are 32,-13 Sum of zeroes =32+-13=9-26=76=-(-7)6=-ba=- Coefficient of x Coefficient of x2
Product of Zeroes =32×-13=-12=ca= Constant term Coefficient of x2
(iv) 4u2+8u
Here, a=4, b=8 and c=0
4u2+8u=4uu+2
Equating this equal to 0 will find values of 2 zeroes of this polynomial.
⇒4uu+2=0
⇒u=0,-2
Therefore, two zeroes of this polynomial are 0,-2 Sum of zeroes =0-2=-2=-21×44=-84=-ba=- Coefficient of x Coefficient of x2 Product of Zeroes =0×-2=0=04=ca= Constant term Coefficient of x2
(v) t2-15
Here, a=1, b=0 and c=-15
We have, t2-15 ⇒t2=15
⇒t=±15
Therefore, two zeroes of this polynomial are 15,-15 Sum of zeroes =15+-15=0=01=-ba=- Coefficient of x Coefficient of x2
Product of Zeroes =15×-15=-15=-151=ca= Constant term Coefficient of x2
(vi) 3x2-x-4
Here, a=3, b=-1 and c=-4
We have, 3x2-x-4=3x2-4x+3x-4
=x3x-4+13x-4=3x-4x+1
Equating this equal to 0 will find values of 2 zeroes of this polynomial. ⇒3x-4x+1=0
⇒X=43,-1
Therefore, two zeroes of this polynomial are 43,-1
Sum of zeroes =43+(-1)=4-33=13=-(-1)3?
=-ba=- Coefficient of x Coefficient of x2
Product of Zeroes =43×(-1)=-43=ca= Constant term Coefficient of x2
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