2. Polynomials

Class X

Exercise 2.1

Q.No.1. The graph of y = p(x) are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:

The numbers of zeroes in Fig. (i) = 0

The number of zero in Fig. (ii) = 1

The numbers of zeroes in Fig. (iii) = 3

The numbers of zeroes in Fig (iv) = 2

The numbers of zeroes in Fig. (v) = 4

The numbers of zeroes in Fig. (vi) = 3

Exercise 2.2

Q.No.1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes 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

Solution:

(i)

x² - 2x - 8 = x² - 4x + 2x - 8 {By the method of splitting the middle term}

= x(x - 4) + 2(x - 4) = (x + 2)(x - 4)

So, the value of the polynomial is zero when 

x + 2 = 0  or, x - 4 = 0

x = -2 or x = 4.

So, the zeroes of the given polynomial are -2 and 4.

Sum of zeroes = -2 + 4 = 2

Product of zeroes = -2x4 = -8

Verification

Sum of zeroes = -(Coefficient of x)/ Coefficient of x²

or, Sum of zeroes = -(-2)/1 = 2

Product of zeroes = Constant term/ Coefficient of x²

or, Product of zeroes = -8/1 = -8

Hence, the relationship between the zeroes and coefficients are verified.

(ii)

4s² - 4s + 1 = 4s² - 2s - 2s + 1 {By the method of splitting the middle term}

= 2s(2s - 1) - 1(2s - 1) = (2s - 1)(2s - 1)

So, the value of the polynomial is zero when
2s - 1 = 0 or, 2s - 1 = 0
or, s = 1/2
So, the zeroes of the given polynomial are 1/2 or 1/2
Sum of the zeroes = 1/2 + 1/2 = 1
Product of zeroes = 1/2x1/2 = 1/4


Verification
Sum of zeroes = -(Coefficient of s)/ Coefficient of s²
Hence, the relationship between the zeroes and coefficients are verified.
(iii)
(iv)(v)(vi)Q.No.2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.Solution:Therefore, (ii)Therefore, (iii)Therefore, (iv)Therefore, (v)Therefore, (vi)Therefore, 






or, Sum of zeroes = -(-4)/4 = 1
Product of zeroes = Product of zeroes = Constant term/ Coefficient of s²
or, Product of zeroes = 1/2x1/2 = 1/4


6x² - 3 - 7x = 6x² - 9x + 2x - 3 {By the method of splitting the middle term}
= 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3)
So, the value of the polynomial is zero when
x = -1/3 or x = 3/2
So, the zeroes of the polynomial are 1/3 or 3/2
Sum of the zeroes = -1/3 + 3/2 = (-2 + 9)/6 = 7/6
Product of the zeroes = -1/3x3/2 = -3/6 = -1/2

Verification

Sum of zeroes = -(Coefficient of x)/ Coefficient of x²

or, Sum of zeroes = -(-7)/6 = 7/6

Product of zeroes = Constant term/ Coefficient of x²

or, Product of zeroes = -3/6 = -1/2

Hence, the relationship between the zeroes and coefficients are verified.

4u² + 8u = 4u(u + 2)
So, the value of the polynomial is zero when
u = 0 or, u = -2
So, the zeroes of the polynomial are 0 and -2.
Sum of the zeroes = 0 + (-2) = -2
Product of the zeroes = 0x-2 = 0

Verification

Sum of zeroes = -(Coefficient of u)/ Coefficient of u²

or, Sum of zeroes = -8/4 = -2

Product of zeroes = Constant term/ Coefficient of u²

or, Product of zeroes = 0/4 = 0

Hence, the relationship between the zeroes and coefficients are verified.

t² - 15 = (t + √15)(t - √15)
So, the value of the polynomial is zero when
t = √15 or, t = -√15
So, the zeroes of the polynomial are √15 and -√15.
Sum of the zeroes = -√15 + √15 = 0
Product of the zeroes = -√15x√15 = 15

Verification

Sum of zeroes = -(Coefficient of t)/ Coefficient of t²

or, Sum of zeroes = 0/1 = 0

Product of zeroes = Constant term/ Coefficient of t²

or, Product of zeroes = 15/1 = 15

Hence, the relationship between the zeroes and coefficients are verified.

3x² - x - 4 = 3x² - 4x + 3x - 4 {By the method of splitting the middle term}
= x(3x - 4) + 1(3x - 4) = (x + 1)(3x - 4)
So, the value of the polynomial is zero when
x = -1 or 4/3.
So, zeroes of the polynomial are -1 and 4/3.
Sum of the zeroes = -1 + 4/3 = 1/3
Product of the zeroes = -1x4/3 = -4/3

Verification

Sum of zeroes = -(Coefficient of x)/ Coefficient of x²

or, Sum of zeroes = -(-1)/3 = 1/3

Product of zeroes = Constant term/ Coefficient of x²

or, Product of zeroes = -4/3 = -4/3

Hence, the relationship between the zeroes and coefficients are verified.

(i) 1/4 , -1 (ii) √2, 1/3 (iii) 0, √5 (iv) 1, 1 (v) -1/4, 1/4 (vi) 4, 1

(i)
Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = 1/4 
Product of zeroes = -1

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  4, b = -1 and c = -4
Hence, the required quadratic polynomial
4x² - x - 4

Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = √2
Product of zeroes = 1/3

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  3, b = -3√2 and c = 1
Hence, the required quadratic polynomial

3x² - 3√2x + 1

Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = 0
Product of zeroes = √5

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  1, b = 0 and c = √5 
Hence, the required quadratic polynomial


x²  + √5

Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = 1
Product of zeroes = 1

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  1, b = -1 and c = 1
Hence, the required quadratic polynomial


x²  - x + 1

Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = -1/4
Product of zeroes = 1/4

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  4, b = -1 and c = 1
Hence, the required quadratic polynomial


4x²  + x + 1


Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = 4
Product of zeroes = 1

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  1, b = -4 and c = 1
Hence, the required quadratic polynomial

x²  - 4x + 1

Exercise 2.3

Q.No. 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x³ - 3x² + 5x - 3, g(x) = x² - 2

(ii) p(x) = x⁴ -3x² + 4x +5, g(x) = x² + 1 - x

(iii) p(x) = x⁴ - 5x + 6, g(x) = 2 - x²

Solution:

Degree of p(x) = 3

Degree of g(x) = 2

Polynomial for first step = x³ - 3x² + 5x - 3

Division for first step = (x³ - 3x² + 5x - 3) ÷ (x² - 2)

Therefore, first term of quotient = x

Polynomial for second step =x³ - 3x² + 5x - 3 - x³ + 2x = -3x² + 7x - 3

Division for second step = (-3x² + 7x - 3)÷ (x² - 2)

Second term of quotient = -3

Polynomial for third step =-3x² + 7x - 3 + 3x² - 6 = 7x - 9

Now degree became less than divisor.

So,

Quotient = x - 3

Remainder = 7x - 9

Continue ................ 

7. Triangles Class IX Mathematics

Exercise 7.1

Q.No. 1. In a quadrilateral ABCD, AC = AD and AB bisect <A (see Fig. 7.16). Show that △ABC⩭△ABD. What can you say about BC and BD.

Solution:

Consider about the △ABC and △ABD.

AB = AB {Common}

AC = AD {Given}

<CAB = <DAB {AB bisect <A}

△ABC ⩭ △ABD {Under SAS}

BC = BD.

Q.No. 2. ABCD is a quadrilateral in which AD = BC and <DAB = <CBA (see Fig. 7.17). Prove that

(i) △ABD ⩭ △BAC

(ii) BD = AC

(iii) <ABD = <BAC.

Solution:

(i)

Consider in △ABD and △BAC.

AB = AB {Common}

AD = BC {Given}

<DAB = <CBA {Given}

△ABD ⩭ △BAC. {Under SAS}Proved.

(ii)

BD = AC {△ABD ⩭ △BAC} Proved.

(iii)

<ABD = <BAC {△ABD ⩭ △BAC}Proved.

Q.No. 3. AD and BC are equal perpendicular to a line segment AB(see Fig. 7.18). Show that CD bisect AB.

Solution:

Consider about the △BOC and △AOD.

BC = AD {Given}

<BOC = <AOD {Vertically opposite angles}

<OBC = <OAD { Each right angles}

△BOC ⩭ △AOD {Under ASA}

BO = AO

∴ CD bisect AB Proved.

Q.No. 4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that △ ABC ⩭ △CDA.

Solution:
Consider about the △ABC and △CDA.
<BAC = <DCA {Alternate pair}
AC = AC {Common}
<ABC = <ADC {Opposite angles of a parallelogram}
∴△ABC⩭△CDA. Proved.

Q.No. 5. Line l is the bisector of an angle <A and B is any point on l. BP and BQ are perpendiculars from B to the arms of <A (see Fig. 7.20).
Show that:
(i) △APB⩭△AQB
(ii) BP = BQ or B is equidistant from the arms <A.

Solution:
(i)
Consider about the △APB and △AQB.
AB = AB {Common}
<APB = <AQB {Each right angles}
<PAB = <QAB {l is the bisector of <A}
∴△APB⩭△AQB {Under ASA} Proved.

(ii)
BP = BQ {△APB⩭△AQB} Proved.

Continuous ............................