\u00a0<\/div>\n
Question 3.
<\/strong>If x \u2013 a is a factor of x3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0a + 2a2<\/sup>x + b, then the value of b is<\/strong>
(a) 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(b) 2<\/strong>
(c) 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d) 3
<\/strong>Solution:
<\/strong>\u2234\u00a0 x \u2013 a is a factor of x3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0a + 2a2<\/sup>x + b
Let f(x) = x3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0a + 2a2<\/sup>x+ b
and x \u2013 a = 0, then x = a
f(a) = a3<\/sup>\u00a0\u2013 3a2<\/sup>.a + 2a2<\/sup>.a + b
= a3<\/sup>\u00a0\u2013 3a3<\/sup>\u00a0+ 2a3<\/sup>\u00a0+ b = b
\u2235\u00a0 x \u2013 a is a factor of f(x)
\u2234\u00a0 \u00a0b = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(a)<\/strong><\/p>\nQuestion 4.
<\/strong>If x140<\/sup>\u00a0+ 2x151<\/sup>\u00a0+ k is divisible by x + 1, then the value of k is<\/strong>
(a) 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>
(b) -3<\/strong>
(c) 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d) -2<\/strong>
Solution:
<\/strong>\u2234 x + 1 is a factor of f(x) = x140<\/sup>\u00a0+ 2x151<\/sup>\u00a0+ k
\u2234 The remainder will be zero
Let x + 1 = 0, then x = -1
\u2234\u00a0 f(-1) = (-1)140<\/sup>\u00a0+ 2(-1)151<\/sup>\u00a0+ k
= 1 + 2 x (-1) + k\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 {\u2235\u00a0 140 is even and 151 is odd}
=1-2+k=k-1
\u2235 Remainder = 0
\u2234 k \u2013 1=0\u00a0 \u21d2 k=1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0(a)<\/strong><\/p>\nQuestion 5.
<\/strong>If x + 2 is a factor of x2<\/sup>\u00a0+ mx + 14, then m =<\/strong>
(a) 7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(b) 2\u00a0 \u00a0 \u00a0 \u00a0<\/strong>
(c) 9\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d) 14
<\/strong>Solution:
<\/strong>x + 2 is a factor of(x) = x2<\/sup>\u00a0+ mx + 14
Let x + 2 = 0, then x = -2
f(-2) = (-2)2<\/sup>\u00a0+ m{-2) + 14
= 4 \u2013 2m + 14 = 18 \u2013 2m
\u2234\u00a0 x + 2 is a factor of f(x)
\u2234\u00a0 Remainder = 0
\u21d2\u00a0 18 \u2013 2m = 0
2m = 18\u00a0 \u21d2\u00a0 m =\u00a018<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0= 9\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c)<\/strong><\/p>\nQuestion 6.
If x \u2013 3 is a factor of x2<\/sup>\u00a0\u2013 ax \u2013 15, then a =<\/strong>
(a) -2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(b) 5<\/strong>
(c) -5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d) 3<\/strong>
Solution:
<\/strong>x \u2013 3 is a factor of(x) = x2<\/sup>\u00a0\u2013 ax \u2013 15
Let x \u2013 3 = 0, then x = 3
\u2234 f(3) = (3)2<\/sup>\u00a0\u2013 a(3) \u2013 15
= 9 -3a- 15
= -6 -3a
\u2234 x \u2013 3 is a factor
\u2234\u00a0 Remainder = 0
-6 \u2013 3a = 0 \u21d2\u00a0 3a = -6
\u2234\u00a0 \u00a0a =\u00a0\u2212<\/span>6<\/span><\/span>3<\/span><\/span><\/span><\/span><\/span>\u00a0= -2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0(a)<\/strong><\/p>\nQuestion 7.
<\/strong>If x51<\/sup>\u00a0+ 51 is divided by x + 1, the remainder is<\/strong>
(a) 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(b) 1<\/strong>
(c) 49\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d) 50
<\/strong>Solution:
<\/strong>Left(x) = x51<\/sup>\u00a0+ 51 is divisible by x + 1
Let x+1=0, then x = -1
\u2234\u00a0f(-1) = (-1)51<\/sup>\u00a0+ 51 =-1+51\u00a0 \u00a0(\u2235 power 51 is an odd integer)
= 50\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 (d)<\/strong><\/p>\nQuestion 8.
<\/strong>If x+ 1 is a factor of the polynomial 2x2<\/sup>\u00a0+ kx, then k =<\/strong>
(a) -2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(b) -3<\/strong>
(c)\u00a0 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d)\u00a0 2
<\/strong>Solution:
<\/strong>\u2234 x + 1 is a factor of the polynomial 2x2<\/sup>\u00a0+ kx
Let x+1=0, then x = -1
Now f(x) = 2x2<\/sup>\u00a0+ kx
\u2234 Remainder =f(-1) =\u00a0 0
= 2(-1)2<\/sup>\u00a0+ k(-1)
= 2 x 1 + k x (-1) = 2 \u2013 k
\u2234 x + 1 is a factor of f(x)
\u2234 Remainder = 0
\u2234 2 \u2013 k = 0 \u21d2\u00a0 k = 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (d)<\/strong><\/p>\nQuestion 9.
<\/strong>If x + a is a factor of x4<\/sup>\u00a0\u2013 a2<\/sup>x2<\/sup>\u00a0+ 3x \u2013 6a, then a =<\/strong>
(a) 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(b) -1<\/strong>
(c) 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d) 2
<\/strong>Solution:
<\/strong>x + a is a factor o f(x) = x4<\/sup>\u2013 a2<\/sup>x2<\/sup>\u00a0+ 3x \u2013 6a
Let x + a = 0, then x = -a
Now, f(-a) \u2013 (-a)4<\/sup>\u00a0-a2<\/sup>(-a)2<\/sup>\u00a0+ 3 (-a) \u2013 6a
= a4<\/sup>-a4<\/sup>-3a-6a = -9a
\u2234 x + a is a factor of f(x)
\u2234Remainder = 0
\u2234 -9a = 0 \u21d2 a = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(a)<\/strong><\/p>\nQuestion 10.
<\/strong>The value of k for which x \u2013 1 is a factor of 4x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0\u2013 4x + k, is<\/strong>
(a) 3<\/strong>
(b) 1<\/strong>
(c) -2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d) -3
<\/strong>Solution:
<\/strong>x- 1 is a factor of f(x) = 4x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0\u2013 4x + k
Let x \u2013 1 = 0, then x = 1
f(1) = 4(1 )3<\/sup>\u00a0+ 3(1)2<\/sup>\u00a0\u2013 4 x 1 + k
= 4+3-4+k=3+k
\u2234 x- 1 is a factor of f(x)
\u2234 Remainder = 0
\u2234 3 + k = 0 \u21d2 k = -3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (d)<\/strong><\/p>\nQuestion 11.
<\/strong>If x+2 and x-1 are the factors of x3<\/sup>+<\/strong>\u00a010x2<\/sup>\u00a0+ mx + n, then the values of m and n are respectively<\/strong>
(a) 5 and-3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(b) 17 and-8<\/strong>
(c) 7 and-18\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d) 23 and -19<\/strong>
Solution:
<\/strong>x+ 2 and x \u2013 1 are the factors of
f(x) = x3<\/sup>\u00a0+ 10x2<\/sup>\u00a0+ mx + n
Let x + 2 = 0, then x = -2
\u2234 f(-2) = (-2)3<\/sup>\u00a0+ 10(-2)2<\/sup>\u00a0+ m(-2) + n
= -8 + 40 \u2013 2m + n = 32 \u2013 2m + n
\u2234\u00a0 x + 2 is a factor of f(x)
\u2234 Remainder = 0
\u2234 32 \u2013 2m + n = 0 \u21d2\u00a0 2m \u2013 n = 32\u00a0\u00a0 \u2026(i)
Again x \u2013 1 is a factor of f(x)
Let x-1=0, then x= 1
\u2234\u00a0 f(1) = (1)3<\/sup>\u00a0+ 10(1)2<\/sup>\u00a0+ m x 1 +n
= 1 + 10+ m + n =m + n+11
\u2234\u00a0 x- 1 is a factor of f(x)
\u2234 m + n+ 11=0 \u21d2 m+n =-11\u00a0 \u00a0 \u00a0 \u00a0\u2026(ii)
Adding (i) and (ii),
3m = 32 \u2013 11 = 21
\u21d2 m =\u00a021<\/span>3<\/span><\/span><\/span><\/span><\/span>\u00a0 = 7
and n = -11 \u2013 m = m = 7, n = -18
\u2234 m= 7, n = -18\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0(c)\u00a0<\/strong><\/p>\nQuestion 12.
<\/strong>Let f(x) be a polynomial such that\u00a0 f(\u00a0\u2212<\/span>1<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0)= 0, then a factor of f(x) is
(a) 2x \u2013 1
(b) 2x + b
(c) x- 1
(d) x + 1
Solution:<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q12.1.png\")
<\/p>\nQuestion 13.
<\/strong>When x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0+ ax \u2013 b is divided by x2<\/sup>\u00a0\u2013 2x-3, the remainder is x \u2013 6. The value of a and b are respectively.<\/strong>
(a) -2, -6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(b) 2 and -6<\/strong>
(c) -2 and 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(d) 2 and 6
<\/strong>Solution:
<\/strong>Let f(x) = x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0+ ax \u2013 b
and Dividing f(x) by x2<\/sup>\u00a0\u2013 2x + 3
Remainder = x \u2013 6
Let p(x) = x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0+ ax \u2013 b \u2013 (x \u2013 6) or x
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-MCQS.jpg\")
<\/p>\n