POLYNOMIALS CLASS 9TH TO 12TH CBSE BOARD, Assignments of Mathematics

Download POLYNOMIALS CLASS 9TH TO 12TH CBSE BOARD and more Assignments Mathematics in PDF only on Docsity! Page |1 2 | POLYNOMIALS EXERCISE 2.1 Q.1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. @ 4x2 — 8x +7 Gi) y2+V2 (ii) BVE+ END (iv) yr 2 (v) x1 + y9 + 150 y Sol. (i) Polynomial in one variable, x Ans. (ii) Polynomial in one variable, y. Ans. Gii) 3Vé+tV2 is not a polynomial as power of ¢t in Vt is not a whole number. Ans. (iv) ye is not a polynomial as power of y in second term, i.e., i =yt y is not a whole number. Ans. (v) x10 + y3 + £59 is not a polynomial in one variable but a polynomial in three variables x, y and ¢. Ans. Q.2. Write the coefficients of x2 in each of the following : D2+P2 ex WZ +x (iii) Fx? +x Civ) VBx-1 Sol. (i) In 2 + x2 + x, coefficient of x? is 1. Ans. (ii) In 2 — x? + x3, coefficient of x? is —1. Ans. (iii) oe +x, coefficient of x? is 5 Ans. (iv) V2x-1, x? is not present hence no coefficient. Ans. Q.3. Give one example each of a binomial of degree 35, and of a monomial of degree 100. Sol. x + 5 is a binomial of degree 35. 2y100 is a monomial of degree 100. Ans. Q.4. Write the degree of each of the following polynomials : @ 5x3 + 4x2 + Tx (ii) 4—y? iii) 5t-V7 (iv) 3 Sol. (i) Degree is 3 as x? is the highest power. Ans. (ii) Degree is 2 as y? is the highest power. Ans. (iii) Degree is 1 as ¢ is the highest power. Ans. (iv) Degree is 0 as x° is the highest power. Ans. Q.5. Classify the following as linear, quadratic and cubic polynomials : @ x2 +x iW) x — x3 (iii) y + yy? + 4 (iv) 1 +x (v) 3t (vi) r? (vii) 7x3 Sol. (i) x? + x is quadratic. (ii) x — x3 is cubic. (iii) y + y? + 4 is quadratic. (iv) 1 + x is linear. (v) 3¢ is linear. (vi) r? is quadratic. (vii) 7x? is cubic.  Page |1 2 | POLYNOMIALS Q.1. Sol. Q.2. Sol. Q3. EXERCISE 2.2 Find the value of the polynomial 5x — 4x? + 3 at @x=0 (ii) x =-1 (iii) x = 2 p(x) = 5x — 4x2 + 3 (i) At x = 0, pO) =5 x 0-4 x 0? + 3 = 3 Ans. (ii) At x = -1, p(-l) = 5 x (-1)-4 x (1)? +3=-5-44+3 =-6 Ans. (iii) At x = 2, p(2)=5 x 2-4 x (22 +3=10-16+3=-3 Ans. Find p(0), p(1) and p(2) for each of the following polynomials : @ py)=y?-y+1 (i) pH=2+t+ 22-8 (iii) p(x) = 33 (iv) px) = («x -1) @ +1) @ po) =y2-y +1 pO) =02-04+1=1 pd) =12-14+1=1 p(2) = 22-2+4+1-=3. Ans. Gi) p® =2+t+ 27-8 pO) =24+04+2x 02-08 =2 pd)=24+14+2x2-B2=4 p(2)=2+24+2x 2?-=44+8-8=4, (iii) p(x) = x3 pO) = 0 pdy=1 p(2) = 8. Ans. (iv) p(x) = (« — D & + 1) pO) = (1) () = -1 pil) = (1-1) (1+ 1) =0 p(2) = (2-1) (2+ 1) =3 Ans. Verify whether the following are zeroes of the polynomial, indicated against them. @ p(x) = 8x + 1, x =-5 (ii) p(x) = 5x — m1, x = : (iii) p(x) = x? - 1, x = 1, -1 (iv) p(x) = (w + 1) @ - 2), x =-1, 2 (v) p(x) = x, x = 0 (vi) p(x) = Ix +m, x = + (vii) p(x) = 8x2 - 1,2 = 1 2 3° V3 (iii) ple) = 2x + 1, x = ;  Page |2 Q.2. Find the remainder when x* — ax? + 6x — a is divided by x — a. Sol. p(x) = x8 - ax? + 6x - a When p(x) is divided by x - a, the remainder is p(a). Substitute x = a in p(x) p(a) = a? -a® + 6a-—a=5a Ans. Q.3. Check whether 7 + 3x is a factor of 3x3 + Tx. Sol. 7+ 3x =0 > 8&=-7 -7 > “x= 3 Substitute « = + in p(x) = 3x3 + Tx p(=2) = of 2) ga(ct) = 34349 = 848-147 __ -490 3 BITS 9 38 9 9 So, remainder = = which is different from 0. Therefore, (8x + 7) is not a factor of the polynomial 3x3 + 7x. Ans.  Page |1 2 | POLYNOMIALS EXERCISE 2.4 Q.1. Determine which of the following polynomials has (x + 1) a factor : @xP+x% 4x41 Wi) xt + x8 4+ x? +x 41 (iii) x4 + 8x3 + 8x2 +041 (iv) x3 — x? — +2 yc + V2 Sol. To have (x + 1) as a factor, substituting x = —1 must give p(-1) = 0. @ xe 4+x2 4x41 = (-1)3 + 1)? + (-1)+1=-14+1-14+1=0 Therefore, x + 1 is a factor of x° + x2 +x +1 Ans. Gi) xt +8 +2 +441 = (-14 + C18 + (1? +(-D+151-14+1-14+1=1 Remainder is not 0. Therefore (x + 1) is not its factor. Ans. iii) x4 + 8x3 + 8x2? +e 41 = (Dt + 8-18 + 8-1)? + (1) +1 =1-3+3-1+1= 1. Remaidner is not 0 Therefore, (x + 1) is not its factor. Ans. (iv) 23 — x? — +254 V2 = (18 - -D? — @+V2y CD + V2 =-1-14 24+ 2+ V2 = 242 Remainder not 0, therefore (x + 1) is not a factor. Ans. Q.2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases : (i) p(x) = 2x3 + x? — 2x -— 1, g(x) =x +1 (ii) p(x) = x3 + 8x2 + 8x +1, g(x) =x + 2 iii) p(x) = x8 — 4x? + x + 6, g(x) =x -— 38 Sol. Gi) g(x) = x + 1. x = — 1 to be substituted in plx) = 2x3 + x2 — 2x —1 p(-1) = 2(-1)8 + (1)? - 2-1) - 1 =-24+14+2-1=0. So, g(x) is a factor of p(x). Ans. (ii) g(x) = x + 2, substitute x = — 2 in p(x) p(x) = x3 + 8x2 + 8x +1 p(-2) = (-2)8 + 3(-2)? + 3-2) + 1=-8+12-64+1=-1. So, g(x) is not a factor of p(x) Ans. Gii) g(x) = x — 3 substitute x = 3 in (x). p(x) = «3 — 4x2 +446 p(3) = (3)8 - 48)? + 3+ 6 = 27 - 36+3+6=0. Therefore, g(x) is a factor of x? — 4x2 + x + 6. Ans.  Page |2 Q.3. Find the value of k, if x — 1 is a factor of p(x) in each of the following cases : WM p@) =x? +x+k Gi) px) = 2x? + kx + V2 iii) p(x) = kx® — J2x4+1 (iv) p(@&) = kx? — 8x +k Sol. (x — 1) is a factor, so we substitute x = 1 in each case and solve for k by making p(1) equal to 0. @ px) =x? +x4+k pd) =1+1+k=0>5k=-2 Ans. Gi) p(x) = 2x? + hx + V2 pd) =2x 12+kx1+ V2 =0 => 2+k+ V2 =0 => k=-2-V2 =-(2+ V2) Ans. (ii) p(x) = kx? - V2x 41 pl) =k- V2 +1=0 = k= 2-1 Ans. (iv) p(x) = kx? — 8x +k pd) =k-3+k=0 >2k-3=0 > ke 3 Ans. Q.4. Factorise : @ 12x — Tx +1 (ii) 2x? + Tx + 3 (iii) 6x? + 5x — 6 (iv) Bx? —x — 4 Sol. (i) 12x? — 7x +1 = 12x? — 4x — 8x +1 = 4x (38x — 1) — 1 (8x - 1) = (4% - 1) (8x — 1) Ans. (ii) 2x? + Tx + 38 = Qe? + 6x +2443 = Qx (x + 8) + 1 (x + 8) = (2x + 1) (& + 8) Ans. (iii) 6x? + 5x - 6 = 6x? + 9x — 4x - 6 = 8x (2x + 3) — 2 (2x + 3) = (8x — 2) (2x + 3) Ans. (iv) 8x? x — 4 = Bx? — dx + 8x — 4 = x (Bx — 4) + 1 (Bx — 4) = (@& + 1) (Bx — 4) Ans. Q.5. Factorise : @ x3 — 2x? —x + 2 (ii) x3 — Bx? — 9x — 5 (iii) x3 + 18x? + 32x + 20 (iv) Qy? + y? — Qy -—1 Sol. (i) p(x) x8 — 2x? -x + 2 Let us guess a factor (x — a) and choose value of a arbitrarily as 1. Now, putting this value in p(x). 1-2-1+2=0 So @ — 1) is a factor of p(x) Now, x? — 2x2 —-x% +2 = x3 —x2— x24 x%-2n +2 =x? (x-D)-x@-1)-2@-1) = (x — 1) (x? — x — 2) = (x — 1) (x? — 2x + x — 2) = (x — 1) (x@ - 2) + 1 @ — 2)} = (« — 1) @ + 1) (&« — 2) Ans. To factorise it x? — MWe +u—-2=x (e—-2)+1@-2)= («+ 1) @- 2). After factorisation (x — 1) (x + 1) (« - 2).  Page |2 Q.3. Factorise the following using appropriate identities : 2 (i) 9x2 + Oxy + yy? (ii) 4y2-4y +1 (iid) ea Sol. (i) 9x? + 6xy + y? = (8x)? + 2 (3x) y + (y)? = (8x + y)? [Using a? + 2ab + b? = (a + b)?] = (8x + y) (8x + y) Ans. (ii) 4y2 — 4y +1 = (2y)? — 2(2y) () + (1)? = (2y — 1)? = (Qy — 1) Qy - 1) (Using a? — 2ab + b? = (a — b)?] Ans. 2 2 (ii) x? 2 = x? (2 1 10 = NZ) [Using a? - b2 = (a +b) (a - 8)] Ans. Q.4. Expand each of the following, using suitable identities : @ (x + Qy + 42)? (ii) (Qn —y +z)? ~— (tii) (2x + By + 22)? 2 (iv) (3a — Tb - c)? (v) (-2x + By — 82)? (vi) [Fe-Z2+1] Sol. ) (x + Qy + 42)? =x? + (Qy)?? + (4z)? + Qn x By + 2x DW x4e+Q2xdexax = x2 + dy? + 1622 + dxy + 16yz + 82x Ans. Gi) (Qx —y + z)? = (2x)? + (Hy) + (2)? + 2 x (2x) Cy) + 2 (-y) (2) + 2 (2) x 2x = 4x2 + y? + 22 — day — Qyz + dex Ans. (iii) (2x + By + 2z)? = (2x)? + (By)? + (2z)? + 2 (2x) (By) + 2 (By) (2z) + 2 (2z) (-2x) = 4x? + Qy? + 422 — 12xy + 12yz — 82x Ans. (iv) (3a — 7b — c)? = (8a)? + (-7b)? + (-c)? + 2 (8a) (-7b) + 2 (-7b) (— c) + 2 (-c) (3a) = 9a? + 49b? + c? — 42ab + 14be — 6ac Ans. (v) (2x + By — 82)? = (2x)? + (Gy)? + 32)? + 2 (2x) (By) + 2 (5y) (-8z) + 2 (-3z) (2x) = 4x? + Q5y? + 922 — 20xy — 30yz + 122% Ans.  Page |3 . 1 1 Poa VY (-1,Y , 2,5(1,)(-1 (vi) [Fo-5e+2 = (52) +(Z] +() +(54\($] +2 Zh pessonge = +4241 Te +1-5a b-b+ta 16 4 2 Q.5. Factorise : @) 4x? + Qy? + 1622 + 12xy — 2dyz — 16xz (ii) Qn? + y? + 822 — 2V2 xy + 4V2 yz — Bxz Sol. (i) 4x? + Qy? + 1622 + 12xy — 24yz — 16xz = (2x)? + (By)? + (4z)? + 2 (2x) (By) + 2(8y) (-4z) + 2 (42) (2x) = (Qx + By — 42)? = (2x + By — 42) (2x + By — 4z) Ans. (ii) Qn? + y? + 822 — 2V2 xy + 42 ye — Bxz = (V2 x)? + (y)? + 22 2)? + 2 (V2 x) Hy) + 2 Cy) C2V22) +2 (V2 x) - 2V2 2) = (V2x-y-— 222) = (W2x -— y — 2V2z) (V2x —y —2V2z) Ans. Q.6. Write the following cubes in expanded form : 3 3 @ (2x +18 (ii) (2a — 36)? (iii) [21] (iv) [»-29| Sol. (i) (2x + 138 = (2x)3 + 13 + 82x) (1) (Qe + D = 8x3 + 1 + 6x (2x + 1) = 8x3 + 12x? + 6x + 1 Ans. (ii) (2a — 3b)8 = (2a)8 — (3b)3 — 3 x 2a x 3b (2a — 3b) = 8a? — 27b3 — 18ab (2a — 3b) = 8a3 — 27b3 — 36a2b + 54ab? Ans. (3x) 2 9f§ =\a(5 «1 = 27 2 rte5e (Fee) 3 Gii) [3« + 1] 8 2 a er re 8 Ans. 8 4 2  . 27 3 (2. 2 2 wo [3] - #-(32) 9@(5> [+39] 3 8 3 2 —-2 yi _2 4 x a7” XY | x 3° 8 4 3 3 2 2 = x° —-— yy? —2x*y+— xy Ans. 27 3 Q.7. Evaluate the following using suitable identities : @ (99)8 Gi) (102)8 Giii) (998) Sol. (@ (99)% 1000000 — 1 — 300 (100 — 1) Gi) (102)8 1000000 + 8 + 600 (100 + 2) 1000000 — 1 — 30000 + 300 = 970299 (100 + 2)? = 100° + 23 + 3(100) (2) (100 + 2) 1000000 + 8 + 60000 + 1200 = 1061208 Ans. Page |4 (100 — 1)® = (100)? + (—1)8 + 3(100) (—1) (100 — 1) Gii) (998)? = (1000 — 2)3 = (1000)? + (—2)3 + 3(1000) (—-2) (998) = (1000)? — 8 — 6000 (998) = 1000000000 — 8 — 5988000 = 994011992 Ans. Q.8. Factorise each of the following : @ 8a? + 63 + 12a2b + 6ab? (ii) 8a® — b® — 12a7b + Gab? (iii) 27 — 125a2 — 135a + 225a? (iv) 64a3 — 27b3 — 144a2b + 108ab2 1 9 1 (ve) 27p®? -——- =p? += P 316 2” 4? Sol. Gi) 8a? + 63 + 12a2b + Gab? = (2a)? + b3 + 3 (2a) (b) (2a + b) = (2a + b)® = (2a + b) (2a + b) (2a + 6) Ans. Gi) 8a3 — 6? — 12a2b + 6ab2 = (2a)? + (-b)? + 3 (2a) (-b) (2a — b) = (2a — b)8 = (2a — b) (2a — b) (2a — b) Ans. Gii) 27 — 125a3 — 135a + 225a? 33 + (5a)? + 3 x (3) (Sa) (3 — 5a) = (3 — 5a)? = (8 — 5a) (3 — 5a) (3 — 5a) Ans. (iv) 64a3 — 2763 — 144a2b + 108ab? = (4a)? + (-3b)3 + 3 (4a) x (-3b) (4a — 3b) = (4a — 3b)® = (4a — 3b) (4a — 3b) (4a — 3d) s_1 92,1 (v) 27) 31672? +4P ll I wo a v v | “eo | + Sey e | HI | —% a wo + v wo | ~ ale g VY | 2 ale < YY 7 /~ Dlr w VY 8 | wi Ole y VY | le Ye Ans.