Page 1 :
Mathematics, (www.tiwariacademy.com), (Chapter โ 2)(Polynomials), (Class - 9), Exercise 2.2, , Question 1:, Find the value of the polynomial 5x โ 4x? + 3 at:, (i) x =0 (ii) x =-1 (iii) x = 2, , โฌ. Answer 1:, , Let p(x) = 5x โ 4x? +3, , (i) Putting x = 0, we get, , p(0) =5x0-4(0)? +3 =3, , (ii) Putting x = โ1, we get, , p(โ-1) =5 x (-1) -4(-1)?2 +3 =-5 - 443 =-6, (iii) Putting x = 2, we get, , p(2) =5x2-โ4(2)?+3=10โ-164+3=-3, , Question 2:, , Find p(0), p(1) and p(2) for each of the following polynomials:, , () pO) =y?-y+1 (i) p(t) =2+ยข4 207-23, (iii) p() = x (iv) p@) = @-DE+D), โฌ. Answer 2:, , (pO) =y? -y+1, , Putting y = 0, we get, p(0) = 0?-0+1=1, , Putting y = 1, we get, pl) = =1+1=1, , Putting y = 2, we get, p(2) =2?-24+1=3, , (ii) p(t) = 24+ยข4 20? - 3, , Putting t = 0, we get, , p(0) =2+0+42(0)?- (0)? =2, , Putting t = 1, we get, , pl) = 24+14+2(0)?-(@)?=24+14+2-1=4, Putting t = 2, we get, , p(2) = 2424 2(2)?-(2)3 =24+24+8-8=4, , www.tiwariacademy.com, A Step towards free Education, , , , 1
Page 2 :
Mathematics, (www.tiwariacademy.com), (Chapter โ 2)(Polynomials), (Class - 9), (iii) p(x) = x?, , Putting x = 0, we get, p(0) = (0)? =0, , Putting x = 1, we get, pl) = (FP =1, , Putting x = 2, we get, p(2) = (2)? =8, , (iv) p@) = -)@+), , Putting x = 0, we get, p(0) = (0-1)O+)=-1, , Putting x = 1, we get, , ppโ) = (4-11 +1) =0x2=0, Putting x = 2, we get, , p22) =(2-Y2+)=3, , Question 3:, Verify whether the following are zeroes of the polynomial, indicated against them., , (i) p@&) = 3x +1; x=-3 (ii) p(@) = 5x โ7; x=, , (iii) px) =x? -1; x=1,-1 (iv) p(x) = (ยซ+ D(-2); x =-1,2, (v) p(x) =x?; x=0 (vi) p@) =lk+m, x=-%, , (vii) p(x) = 3x? -1; x= 7 s (viii) p(x) = 2x +1; x=5, , โฌ. Answer 3:, , (D) p@) = 3x41; x= -3, , Putting x = -} we get, , ( 3) =3x( s)+1= 1+1=0, , P\3) = 3) ~, , Here, p (- >) = 0, Hence, x = -: is a solution of p(x) = 3x +1., (ii) p(x) = 5x -โ 7; x=, , Putting x = : we get, , rf) =sx(@)โr=4, Here, p (2) # 0, Hence, x = : is not a solution of p(x) = 5x โ 1., , www.tiwariacademy.com, A Step towards free Education, , , , 2
Page 3 :
Mathematics, (www.tiwariacademy.com), (Chapter โ 2)(Polynomials), (Class - 9), , (iii) p(x) =x? -1; x=1,-1, , Putting x = 1, we get, , p(1) = (1)?-1=1-1=0, , Here, p(1) = 0, Hence, x = 1 isasolution of p(x) = x? โ1., , Putting x = โ1, we get, , p(-1) = (-1)??-1=1-1=0, , Here, p(โ1) = 0, Hence, x = โ1 isasolution of p(x) = x? โ1., , (iv) p@&) = (ยซ+ 1)(x- 2); x =-1,2, , Putting x = โ1, we get, , p(-1) = (-1+ 1)(-1 - 2) = 0x (-3) =0, , Here, p(โ1) = 0, Hence, x = โ1 is asolution of p(x) = (x + 1)(x โ 2)., , Putting x = 2, we get, , p(2) = (24+ 1)(2-2) =3x9=0, Here, p(2) = 0, Hence, x = 2 is a solution of p(x) = (x + 1)(x โ 2)., , (v) p(x) =x?; x=0, , Putting x = 0, we get, , p(0) = (0)? =0, , Here, p(0) = 0, Hence, x = 0 is a solution of p(x) = x?., , (vi) p(x) = lx +m; x= โ=, Putting x = โ" we get, m m, p(-7)=!x(-7)+m=-mtm=0, Here, p (โโข) = 0, Hence, x = โisa solution of p(x) =lx +m., , (vii) p(x) = 3x2 -1; x= -= s, Putting x = โ, we get, 1 1)? 1, p(-=)=3(-s) w~E=3xg-1=1-L=0, Here, p (- =) = 0, Hence, x = -zis a solution of p(x) = 3x? -1., , Putting x = = we get, 2, , r(=)-3(4) -1=3xg-1=4-153, , Here, p (=) # 0, Hence, x = zis nota solution of p(x) = 3x? -1., , www.tiwariacademy.com, A Step towards free Education, , , , 3
Page 4 :
Mathematics, , (www.tiwariacademy.com), (Chapter โ 2)(Polynomials), (Class - 9), (viii) p(x) = 2x 41; x =5, Putting x = > we get, , (5) =2x(5)+1s1+1=2, PUQ) eg) Tee ar eยฎ, Here, p G) # 0, Hence, x = zis not a solution of p(x) = 2x + 1., , Question 4:, Find the zero of the polynomial in each of the following cases:, , () p@&) =x4+5 (ii) p&) =x-5 (iii) p(x) = 2x +5, (iv) p(x) = 3x -2 (v) p(x) = 3x (vi) p(x) = ax, a #0, (vii) p(x) = cx +d; c #0, c, dare real numbers., , โฌ. Answer 4:, , (i) p@) =x4+5, Putting p(x) = 0, we get, , x+5=0, =>x=-5, Hence, x = โ5 is a zero of the polynomial p(x)., , (ii) p@) =x-5, Putting p(x) = 0, we get, , x-5=0, , =>x=5, , Hence, x = 5 is a zero of the polynomial p(x)., , (iii) p(x) = 2x +5, Putting p(x) = 0, we get, , 2x+5=0, Sx=-t, 2, Hence, x = -3 is a zero of the polynomial p(x)., , (iv) p(x) = 3x -2, Putting p(x) = 0, we get, 3x-2=0, , 2, Sx=t, 3, , Hence, x = : is a zero of the polynomial p(x)., , (v) p@) = 3x, , Putting p(x) = 0, we get, 3x =0, , >x=0, , Hence, x = 0 is a zero of the polynomial p(x)., , www.tiwariacademy.com, A Step towards free Education, , , , 4
Page 5 :
Mathematics, (www.tiwariacademy.com), (Chapter โ 2)(Polynomials), (Class - 9), , (vi) p(x) = ax, a#0, , Putting p(x) = 0, we get, , ax =0, , >xk=0, , Hence, x = 0 is a zero of the polynomial p(x)., , (vii) p(x) = cx +d; c #0, c, dare real numbers., Putting p(x) = 0, we get, , cx+d=0, a, pxs-a, fa, Hence, x = โ~ isa zero of the polynomial p(x)., , www.tiwariacademy.com, A Step towards free Education, , , , 5
