[PDF] Important-Questions-ICSE-Class-10th-Maths-Factor-Theorem-Year

101381_6Important_Questions_ICSE_Class_10th_Maths_Factor_Theorem_Year_2009.pdf Important Questions ICSE Class 10th : Maths Year 2009 ( F a c t o r T h e o re m) Q u e s t i o n 1 . Sh o w t h a t ( x - 1 ) i s a f a c t o r o f x

3 - 7x2 + 14x - 8. Hence, factorize the above

p o l y n o m i a l c o m p l e t e l y ? S o l u t i o n : I f ( x - 1 ) i s a f ac t o r o f f ( x ) = x3 - 7x2 + 14x - 8, then f(1) = 0. N o w , f ( 1 ) = 1

3 - 7(1)2 + 14(1) - 8 = 1 - 7 + 14 - 8 = 0.

H e n ce , ( x - 1 ) i s a f ac t o r o f f ( x ). T o f i n d o t h e r f ac t o r , f ( x ) = x

3 - 7x2 + 14x - 8 = x2(x - 1) - 6(x - 1) + 8(x - 1)

= ( x - 1 ) (x

2 - 6x +8) = (x - 1){x(x - 4) - 2(x - 4)}

= ( x - 1 )( x - 2 )( x - 4 ) . [ A n s .] Q u e s t i o n 2 . F i nd t h e r e m a i nd e r w h e n f ( x ) = x

3 - 6x2 + 9x + 7 is divided by g(x) = x - 1 ?

S o l u t i o n : W h e n f ( x ) i s d i v i d e d b y g ( x ) = x - 1 , t h e n r e m a i n d e r , R = f ( 1 ) , b y r e m a i n d e r t h e o r e m. H e n ce , R = f ( 1 ) = ( 1)

3 - 6 (1)2+9(1) + 7

= 1 - 6 + 9 + 7 = 11 . [ A n s .] Q u e s t i o n 3 . F o r w h a t va l u e o f ' a ' , t h e p o l y n o m i a l g ( x ) = x - a i s a f a c t o r o f f ( x ) = x

3 - ax2 +x + 2

? S o l u t i o n : A s , x - a i s a f ac t o r o f f ( x ) , t h e r e f o r e , f ( a ) = 0 i . e . a

3 - a × a2 + a + 2 = 0

o r , a

3 - a3 + a + 2 = 0

o r , a + 2 = 0 o r , a = - 2 . [ A n s .] Q u e s t i o n 4 . F i nd t h e va l u e o f p a nd q , i f ( x + 3 ) a nd ( x - 4) are the factors of x3 - px2 - qx + 24 ? S o l u t i o n : L e f ( x ) = x3- px2 - qx + 24. A s , x + 3 i s a f ac t o r o f f ( x ) , f ( - 3 ) = 0 i . e . ( - 3)

3 - p(- 3)2 - q(- 3) + 24 = 0

o r , - 27
- 9p + 3q + 24
= 0 h ttp:// www . c bs e p o r t a l . c om or, - 9p + 3q - 3 = 0 --------------------------- (i) A l s o x - 4 i s a f ac t o r o f f ( x ) , t h e n f ( 4 ) = 0 i . e . ( 4 )

3 - p(4)2 - q(4) + 24 = 0

o r , 64
- 16p - 4q + 24
= 0 o r , - 16p - 4q + 88
= 0 ----------------------( ii) M u l t i p l y i n g e q n . ( i ) b y 4 a n d e q n . ( ii ) b y 3 a n d t h e n a dd i n g w e g et - 84p
+ 252
= 0 o r , p = 3 . [ A n s .] P u tt i n g v a l u e o f p i n e q n . ( i ) w e g e t, 9 × 3 + 3q - 3 =0 O r , 3q - 30
= 0 O r , q = 10 . [ A n s .] Q u e s t i o n 5 . F i nd t h e va l u e o f q i f t h e p o l y n o m i a l f(x) = x3 + 2x2 - 13x + q is divisible by g(x) = x - 2 . H e n ce f i nd a ll t h e f a c t o r s ? S o l u t i o n : A s , f ( x ) = x3 + 2x2 - 13x + q is divisible by g(x) = x - 2, T h e r e f o r e , f ( 2 ) = 0 O r , ( 2)

3 + 2(2)2 - 13(2) + q = 0

O r , 8 + 8 - 26
+ q = 0 O r , q = 10 . [ A n s .] N o w , f ( x ) = x

3 + 2x2 - 13x + 10

= x

3 - 2x2 + 4x2 - 8x - 5x + 10

= x

2(x - 2) + 4x(x - 2) - 5(x - 2)

= ( x - 2 )(x

2 + 4x - 5)

= ( x - 2 )(x

2 + 5x - x - 5)

= ( x - 2 ) { x ( x + 5 ) - 1 ( x + 5 )} = ( x - 2 )( x - 1 )( x + 5 ) . [ A n s .] h ttp:// www . c bs e p o r t a l . c om