Ncert Solutions For Class 10 Maths Chapter 289
Ncert Solutions For Class 10 Maths Chapter 289
Ncert Solutions For Class 10 Maths Chapter 289
Solutions:
(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis
does not cut it at any point.
(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at
only one point.
(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at
any three points.
(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at
two points.
(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at
four points.
(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at
three points.
NCERT Solution For Class 10 Maths Chapter 2- Polynomials
Solutions:
(i)x2–2x –8
⇒x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)
(ii)4s2–4s+1
⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)
s2 )
(iii) 6x2–3–7x
⇒6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x - 3) +1(2x - 3) = (3x+1)(2x-3)
Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2)
(iv)4u2+8u
⇒ 4u(u+2)
(v) t2–15
⇒ t2 = 15 or t = ±√15
Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)
(vi) 3x2–x–4
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes
respectively.
(i) 1/4 , -1
Solution:
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2–(1/4)x +(-1) = 0
4x2–x-4 = 0
(ii)√2, 1/3
Solution:
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2 –(√2)x + (1/3) = 0
3x2-3√2x+1 = 0
(iii) 0, √5
Solution:
Given,
Sum of zeroes = α+β = 0
Product of zeroes = α β = √5
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2–(0)x +√5= 0
(iv) 1, 1
Solution:
Given,
Sum of zeroes = α+β = 1
Product of zeroes = α β = 1
NCERT Solution For Class 10 Maths Chapter 2- Polynomials
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2–x+1 = 0
Solution:
Given,
Sum of zeroes = α+β = -1/4
Product of zeroes = α β = 1/4
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2–(-1/4)x +(1/4) = 0
4x2+x+1 = 0
(vi) 4, 1
Solution:
Given,
Sum of zeroes = α+β = 4
Product of zeroes = αβ = 1
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly
as:-
x2–(α+β)x+αβ = 0
x2–4x+1 = 0
Solution:
Given,
Dividend = p(x) = x3-3x2+5x–3
Divisor = g(x) = x2– 2
Solution:
Given,
Dividend = p(x) = x4 - 3x2 + 4x +5
Divisor = g(x) = x2 +1-x
NCERT Solution For Class 10 Maths Chapter 2- Polynomials
Remainder = -5x + 10
2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial
by the first polynomial:
As we can see, the remainder is left as 0. Therefore, we say that, t2-3 is a factor of 2t4 +3t3-2t2 -9t-12 .
(ii)x2+3x+1 , 3x4+5x3-7x2+2x+2
Solutions:
Given,
As we can see, the remainder is left as 0. Therefore, we say that, x2 + 3x + 1 is a factor of 3x4+5x3-7x2+2x+2.
Solutions:
Given,
As we can see, the remainder is not equal to 0. Therefore, we say that, x3-3x+1 is not a factor of x5-4x3+x2+3x+1 .
NCERT Solution For Class 10 Maths Chapter 2- Polynomials
3. Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeroes are √(5/3) and - √(5/3).
Solutions:
Since this is a polynomial equation of degree 4, hence there will be total 4 roots.
Now, when we will divide f(x) by (3x2−5) the quotient obtained will also be a factor of f(x) and the remainder
will be 0.
x2+2x+1 = x2+x+x+1 = 0
x(x+1)+1(x+1) = 0
(x+1)(x+1) = 0
4. On dividing x3-3x2+x+2 by a polynomial g(x), the quotient and remainder were x–2 and –2x+4, respectively.
Find g(x).
Solutions:
Given,
Dividend, p(x) = x3-3x2+x+2
Quotient = x-2
Remainder = –2x+4
We have to find the value of Divisor, g(x) =?
As we know,
Dividend = Divisor × Quotient + Remainder
5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solutions:
According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we
NCERT Solution For Class 10 Maths Chapter 2- Polynomials
can find the value of quotient q(x) and remainder r(x), with the help of below given formula;
Solution:
Given, p(x) = 2x3+x2-5x+2
p(1) = 2(1)3+(1)2-5(1)+2 = 0
p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0
∴ ax3+bx2+cx+d = 2x3+x2-5x+2
α +β+γ = –b/a
αβ+βγ+γα = c/a
α βγ = – d/a.
Hence, the relationship between the zeroes and the coefficients are satisfied.
NCERT Solution For Class 10 Maths Chapter 2- Polynomials
Solution:
Given, p(x) = x3-4x2+5x-2
∴ p(2)= 23-4(2)2+5(2)-2 = 0
∴ ax3+bx2+cx+d = x3-4x2+5x-2
a = 1, b = -4, c = 5 and d = -2
α + β + γ = –b/a
αβ + βγ + γα = c/a
α β γ = – d/a.
Hence, the relationship between the zeroes and the coefficients are satisfied.
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the
product of its zeroes as 2, –7, –14 respectively.
Solution:
Let us consider the cubic polynomial is ax3+bx2+cx+d and the values of the zeroes of the polynomials be α, β, γ.
α βγ = -d/a = -14/1
NCERT Solution For Class 10 Maths Chapter 2- Polynomials
Thus, from above three expressions we get the values of coefficient of polynomial.
a = 1, b = -2, c = -7, d = 14
Solution:
We are given with the polynomial here,
p(x) = x3-3x2+x+1
∴px3+qx2+rx+s = x3-3x2+x+1
p = 1, q = -3, r = 1 and s = 1
Sum of zeroes = a – b + a + a + b
-q/p = 3a
-(-3)/1 = 3a
a=1
-s/p = 1-b2
-1/1 = 1-b2
b2 = 1+1 = 2
b = ±√2
4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 ±√3, find other zeroes.
Solution:
Since this is a polynomial equation of degree 4, hence there will be total 4 roots.
NCERT Solution For Class 10 Maths Chapter 2- Polynomials
∴ [x−(2+√3)] [x−(2-√3)] =0
(x−2−√3)(x−2+√3) = 0
On multiplying the above equation we get,
Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.