1. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).
2. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).
3. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).
4. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).
5. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).
6. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).
7. 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.
8. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
9. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
10. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
(ii) x2 + 3x +1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
- Solution:no zeroes.
2. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).- Solution:One
3. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).- Solution:Three
4. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).- Solution:Two
5. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).- Solution:Four
6. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).- Solution:Three.
7. 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= x(x – 4) + 2(x – 4)= (x – 4)(x + 2)Therefore the zeroes of the polynomial x2 – 2x – 8 are {4, -2}.Relationship between the zeroes and the coefficients of the polynomial:Sum of the zeroes = -= -Also sum of the zeroes of the polynomial = 4 – 2 = 2.Product of the zeroes =
=
Also product of the zeroes = 4 x –2 = -8Hence verified.(ii) 4s2 – 4s + 1= 4s2 – 2s – 2s + 1= 2s(2s – 1) – 1(2s – 1)= (2s – 1)(2s – 1)= 2(s -2(s -
Therefore the zeroes of the polynomial areRelationship between the zeroes and the coefficients of the polynomial:Sum of the zeroes = -= -
= 1
Also sum of the zeroes == 1
Product of the zeroes ==
Also product of the zeroes =Hence verified.(iii) 6x2 – 3 – 7x= 6x2 – 7x – 3= 6x2 – 9x + 2x – 3= 3x(2x – 3) + 1(2x – 3)= (2x –3)(3x + 1)= 2(x -3(x +
= 6(x -(x +
The zeroes of the polynomials are {
Relationship between the zeroes and the coefficients of the polynomial:Sum of the zeroes = -= -
=
Also sum of the zeroes =Product of the zeroes =Also product of the zeroes =Hence verified.(iv) 4u2 + 8u= 4u(u + 2)= 4[u – 0][u –(- 2)]The zeroes of the polynomials are {0, -2}Relationship between the zeroes and the coefficients of the polynomial:Sum of the zeroes = -= -
Also sum of the zeroes =Product of the zeroes ==
Also product of the zeroes =Hence verified.(v) t2 – 15= (t +The zeroes of the polynomials are {Relationship between the zeroes and the coefficients of the polynomial:Sum of the zeroes = -= -
Also sum of the zeroes =Product of the zeroes ==
Also product of the zeroes =Hence verified.(vi) 3x2 – x – 4.= 3x2 – 4x + 3x – 4= x(3x – 4) + 1(3x – 4)= (3x – 4)(x + 1)The zeroes of the polynomials are {Relationship between the zeroes and the coefficients of the polynomial:Sum of the zeroes = -Also sum of the zeroes =Product of the zeroes =Also product of the zeroes =Hence verified.
8. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(ii) 
(iii) 0, 
(iv) 1, 1 (v) -
, - Solution:(i) Let the quadratic polynomial be ax2 + bx + c, and its zeroes be a and bGiven a + b =
=
a b = -1 =If a = 4 , b = -1 and c = -4The quadratic polynomial is 4x2 - x - 4.(ii)Let the quadratic polynomial be ax2 + bx + c, and its zeroes be a and bGiven a + b ==
a b ==
If a = 1, b = -and c =
The quadratic polynomial x2 -x +
(iii) 0,Let the quadratic polynomial be ax2 + bx + c and the zeroes be a + bGiven a + b = 0 =a b ==
If a = 1, b = 0 and c =The quadratic polynomial is x2 +(iv) 1, 1Let the quadratic polynomial be ax2 + bx + c , and its zeroes be a + bGiven a + b == 1
a b == 1
\ If a = 1 , b = -1 and c = 1\ The quadratic polynomial is x2 - x + 1.(v),
Let the quadratic polynomial be ax2 + bx + c , and its zeroes be a + b
Given a + b ==
a b =.
If a = 4, b = 1 and c = 1The quadratic polynomial is 4x2 + x + 1.(vi) 4, 1Let the quadratic polynomial be ax2 + bx + cGiven a + b =a b =If a = 1, b = -4 and c = 1\ The quadratic polynomial is x2 – 4x +1.
9. 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
- Solution:(i) p(x) = x3 – 3x2 + 5x + 3, g(x) = x2 – 2
Quotient is (x - 3)Remainder is 7x – 9(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – xRearrange g(x) as x2 - x + 1The Quotient is x2 + x – 3Remainder is 8(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2Rearrange g(x) as –x2 + 2Quotient is –x2 – 2Remainder is –5x + 10
10. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
(ii) x2 + 3x +1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
- Solution:(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12\ 1st is a factor of the second polynomial.(ii) x2 + 3x+1, 3x4 + 5x3 – 7x2 + 2x + 2
Hence 1st polynomial is a factor of the second polynomial(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
No it is not a solution- Solution:Two zeros are
and -
Since the two zeros are(x -is a factor of the polynomial (i.e.,) (3x2 – 5)
\ Divide the given polynomial by 3x2 – 5.(3x4 + 6x3 – 2x2 – 10x – 5) = (3x2 – 5) (x2 + 2x + 1)= (3x2 – 5) (x + 1)2Its zeroes are given by x = -1, x = -1The zeroes of the given polynomial are
- Solution:By the division algorithm,Dividend = Divisor ´ Quotient + Remainder.In this problem f(x)= x3 – 3x2 + x + 2, since (-2x + 4 ) is the remainder, subtract (-2x + 4) from f(x) then divide the result by (x - 2).
Thus g(x) =x2 – x + 1.
- Solution:If p(x) denotes the dividend, g(x) denotes the divisor, q(x) denotes the quotient, r(x) denotes the remainder.p(x) = 3x2 – 6x + 12 , g(x) = 3 , q(x) = x2 - 2x + 4, r(x) = 0.p(x) = x3 + x2 + x +1, g(x) = x2 – 1, q(x) = x + 1, r(x) = 2x + 2.p(x) = 2x3 + 2x2 – 2x +2, g(x) = x2 – x –1, q(x) = 2x + 4, r(x) = 6.
- Solution:Two zeros are
and -
.
