CG SCERT Chhattisgarh State Board Class 10 Mathematics Chapter 1 Polynomials Exercises Questions and Answers.
Exercise 1
cg polynomials(11) length x breath = area
⇒ length x (15x) = 45x2 + 30x
⇒ length = 45x2/15x + 30x/15x = (3x+2)
(12) 25x into 2 equal parts is 14x
Exercise 2
(1)
(i) p(x) = x3 + 3x2 – 5x + 8
q(x) = x + 1
r = p(-1); now putting x = -1
∴Remainder; p(-1) = (-1)3 + 3 ( – 1)2 – 5 (- 1) + 8
= -1 + 3 + 5 + 8
= 15
(ii)p(x) = x3 + 3x2 – 5x + 8
q(x) = 2x – 1 ; 2[x- 1/2]
r = p (1/2); now putting x = 1/2
∴Remainder; p(1/2) = (1/2)3 + 3 (1/2)2 – 5(1/2) +8
= 1/8 + 3/4 – 5/2 + 8
= 1+6 – 20+64/8
= 51/8
(iii)p(x) = x3 + 3x2 – 5x +8
g(x) = x+2 ;
r = p (-2); now putting x = -2
∴Remainder; p(-2) = (-2)3 + 3 (-2)2 – 5 (-2) + 8
= – 8 + 12 + 10 + 8
= 22
(iv)p(x) = x3 + 3x2 – 5x + 8
g(x)= x – 4
r = p(4); now putting x = 4
∴Remainder; p(4) = (4)3 + 3 (4)2 -5 (4) + 8
= 64 +48 – 20 + 8
= 100
(v) p(x) = x3 + 5x2 – 5x + 8
q(x)= x+1/3;
r = p(-1/3); now putting x = -1/3
∴Remainder; p(-1/3) = (-1/3)3 + 3 (-1/3)2 – 5 (-1/3)+ 8
= -1/27+ 3/9 + 5/3+8
= -1/27 + 1/3 + 5/5 + 8
= -1+ 9+45+216/27
= 269/27
(2)
(i) p(x) = x3 – 4x2 + x = 6
q(x) = x – 3
r = p(3); now putting x = 3
∴Remainder; p(3) = (3)3 – 4 (3)2 + (3) + 6
= 27- 36 + 3+ 6
= 30-30
= 0
∴ (x-3) is a factor of given polynomial
(ii) p(x) = 2x3 + x2 – 2x + 1
q(x)= x + 1
r = p(-1)
now; p(-1) = 2(-1)3 + (-1)2 – 2(-1) +1
= -2+ 1 +2+1
=2
∴ 9x+1) is not factor of given polynomial
(iii)p(x) = x4– x3 – x2–x-2
g(x) = x-2
r = p(2)
Now, p(2) = (2)4 – (2)3 – (2)2 –2 – 2
= 16 -8 – 4 – 2-2
= 16-16
=0
∴ (x-2) is a factor of given polynomial
(iv)p(x) = x3+ 5x2 – 5x +1
q(x)= x-1
r = p (1) = (1)3 + 5 (1)2 -5 (1) +1
= 1+ 5 –5+ 1
= 2
∴ (x-1) is not a factor of given polynomial
(v)p(y) = y2+2y-1
g(y) = y+ 4
r = p (-4) = (-4)2 + 2(-4) -1
= 16 – 8-1
=7
∴ (y+4) is not a factor of a given polynomial
(3)
(i) p(x) = x2 +ax + 2
q(x) = x+1
r = p (-1);
p(-1) = (-1)2 + a(-1)+2=a
⇒ -a=-3⇒ a = 3
∴ a = 3
(ii)p(x) = ax2 +5x + 3
q(x) = x-1
r = p (1);a(1)2-5(1)+3=0
⇒ -a=2
∴ a = 2
(iii)p(x) = 2x2 +6x+a
g(x)= x+2
r= p(-2); 2(-2) + 6(-2) + a =0
⇒8+ (-120 +a = 0
⇒a = 4
∴ a = 4
(iv)p(t) = t2+2at-2a+3
g(t)= t-3
r= p(3);
(3)2+2a(3)-2a+3=a
⇒9+6a-2a+3=0
⇒4a = -12
⇒a = -3
∴ a = -3
(v)p(y) = y2-2y+a
g(y)= y+5
r= p(-5); (-5)2-2(-5) + a = 0
⇒25+10+a=0
⇒a = -35
(4) f(x) = (x2-9) + (3x+2)
⇒f(x)/(x-3) = (x+3) (x-3)/(x-3) q(x) + 3x+2/x-3
Remainder = remainder when (3x+2) is divided by (x-3)
=3 X (5)+2
= 11
(5) f(x) = (x2-16)+(5x+3)
⇒f(x)/(x+4) = (x+40 (x-4)/(x+4) q(x) + 5x+3/x+4
Remainder = Remainder when (5x+3) is divided by (x+4)
= 5(-4)+3
= -17
Exercise 3
(1)x2-3x-4 = x2-4x+x-4 = x (x-4)+ 1 (x-4) = (x-4)(x+1)
(2)x2 + 2x +1 = x2 +x+x+1 = x(x+1) + 1 (x+1) = (x+1) (x+1)
(3)x2+x-12 = x2 + 4x-3x-12 = x(x+4) -3 (x+4) = (x+4) (x-3)
(4)x2-8x +15 = x2 – 5x -3x+15 = x(x-5) -3 (x-5) = (x-5) (x-5)
(5)-y2 + 35y + 156 = -y2 + 39y -4y + 156 = -(y-39) (y+4)
(6)t2 -4t- 21 = t2 -7t + 3t -21 = t (t-7) + 3 (t-7) = (t-7) (t+3)
(7)7x2– 2x-5 = 7x2 -7x + 5x-5 = 7x (x-1) +5 (x-1) = (7x+5) (x-1)
(8) 12x2– 24x +12 = 12x2 – 12x – 12x+12 = 12(x-1) (x-1)
(9) 6x2 -7x-3 = 6x2-9x+2x-3 = (2x-3) (3x+1)
(10) 14y2 + 19y-3 = 14y2+21y-21y-2y-3 = 7y (2y+3) -1 (2y+3) = (2y+3) (7y-1)
(11) √3y2 +9y+6√3 = √3y2 + 6y + 6√3y+6√3 = (y+2√3) (√3y+3)
(12) 144x2 +24x+1 = (12x)2 + 2 x 12x x 1 + (1)2 = (12x+1)2
Exercise 4
(1)
(i) (3,4) ⇒ (x-3) (x-4) = x2 -4x -3x+12
=x2– 7x+12
(ii) (-2,-3) ⇒(x+2) (x+3)
(iii) (1/2, -1/2) ⇒(x – 1/2) (x+1/2)
(iv) (15,17) ⇒ (x-15) (x-17)
(v) (-18,12) ⇒(x+18) (x-12)
(2)
(i)x2+10x+24;
Sum = -b/a = -10/1 =-10
Product = c/a = 24/1 = 24
(iv) -5x2 + 3x+4;
Sum = -b/a =-3/(-5) = 3/5
Product = c/a = 4/(-5) = – 4/5
(ii) 2x2-7x-9;
Sum = -b/a = -(-7)/2 = 7/2
Product = c/a = -9/2
(iii)x2+ 11x+30;
Sum = -b/a = -11/1= -11
Product = c/a = 30/1 = 30
(v)x2+x-12;
Sum = -b/a = -1/1 = 1
Product = c/a = -12/1 = -12