Module 10
Factors of Polynomials
WHAT’S MORE ( y−5)( y 2 +5 y +25)
Difference of Cubes
1) 64 x 3 +27
WHAT I HAVE LEARNED
( 4 x )3 + ( 3 )3
( 4 x+ 3)(16 x 2−12 x+ 9) 1) Factor 8 x 3 y 2 z+12 x 2+ y z 2 +16 x y 3 z3
Sum of Cubes
2) Factor c 3−1
2) 25 x 2−70 x+ 49
( 5 x )2−[ ( 2 )( 5 x )( 7 ) ]+ (7 )2 ( c )3− ( 1 )3
(c−1)(c 2 +c +1)
( 5 x−7 )2
Perfect Square Trinomial
3) Factor c 3 +1
3) 8 x 3 y −50 y3 x ( c )3 + ( 1 )3
2 xy ( 4 x 2−25 y 2 ) ( c +1 ) ( c 2−c +1 )
2 xy ( 2 x−5 y )( 2 x +5 y )
Quadratic Trinomial 4) Factor 3 x 2 y 2−12
4) y 3−125 3 ( x 2 y 2−4 )
( y )3−( 5 )3 3 [ ( xy )2−( 2 )2 ]
3 ( xy−2 ) ( xy +2 )
WHAT I CAN DO
1) 2 x2 −3 x −20=( 2 x+ 5 )( x−4 )
2 x2 −3 x −20=[ ( 2 x )( x ) ] + [ ( 2 x ) (−4 ) ]+ [ (5 )( x ) ] + [ ( 5 ) (−4 ) ]
2 x2 −3 x −20=( 2 x 2 ) + (−8 x )+ (5 x ) + (−20 )
2 x2 −3 x −20=2 x 2−8 x+ 5 x−20
2 x2 −3 x −20=2 x 2−3 x−20
∴ ( 2 x +5 )∧( x−4 ) are factorsof 2 x2 −3 x −20
2) 100 x 2−20 xy+ y 2=( 10 x − y )( 10 x + y )
100 x 2−20 xy+ y 2=[ ( 10 x ) ( 10 x ) ] + [ ( 10 x )( y ) ] + [ (− y ) ( 10 x ) ] + [ (− y ) ( y ) ]
100 x 2−20 xy+ y 2=( 100 x2 ) + (10 xy ) + (−10 xy ) + (− y 2 )
100 x 2−20 xy+ y 2=100 x2 +10 xy−10 xy− y 2
100 x 2−20 xy+ y 2 ≠ 100 x 2− y 2
∴ ( 10 x− y )∧( 10 x + y ) are not factors of 100 x 2−20 xy+ y 2
3) b 3−c 3=( b−c ) ( b2 +bc +c 2 )
b 3−c 3={ [ ( b ) ( b 2) ]+ [ ( b ) ( bc ) ] + [ ( b ) ( c 2 ) ] }+ {[ (−c ) ( b2 ) ]+ [(−c )( bc ) ] + [ (−c ) ( c 2 ) ] }
b 3−c 3=[ ( b3 ) + ( b 2 c ) + ( b c 2) ]+ [ (−b 2 c ) + (−b c 2) + (−c 3 ) ]
b 3−c 3=( b 3 +b2 c+ b c 2 ) + (−b 2 c−b c 2−c 3 )
b 3−c 3=b 3+ b2 c+ b c 2−b2 c−b c2−c 2
b 3−c 3=b 3−c 3
∴ ( b−c )∧( b2 +bc +c 2 ) are factors of b3−c3
4) k 3+ 8 m3 =( k +2 m ) ( k 2−2 km+ 4 m2 )
k 3+ 8 m3 ={[ ( k ) ( k 2 ) ] + [ ( k )(−2 km ) ]+ [ ( k ) ( 4 m2) ] }+ { [ ( 2 m) ( k 2) ] + [ ( 2m ) (−2 km ) ] + [ ( 2m ) ( 4 m2 ) ] }
k 3+ 8 m3 =[ ( k 3 ) + ( −2 k 2 m ) + ( 4 k m 2 ) ] + [ ( 2 k 2 m ) + (−4 k m 2 ) + ( 8 m 3 ) ]
k 3+ 8 m3 =( k 3−2k 2 m+4 k m 2) + ( 2 k 2 m−4 k m2 +8 m 3 )
k 3+ 8 m3 =k 3−2 k 2 m+4 k m2 +2 k 2 m−4 k m2+ 8 m3
k 3+ 8 m3 =k 3 +8 m 3
∴ ( k +2 m )∧( k 2−2 km+ 4 m 2 ) are factors of k 3 +8 m 3
ASSESSMENT ( 12 x 2+28 x )− (15 x−35 )
1) B. 4 x ( 3 x+ 7 )−5 ( 3 x+ 7 )
( 4 x−5)(3 x+7)
2) A.
7 x 4 +35 x 3 10) C.
7 x 3 ( x+5) 3 x 3+5 x 2 +9 x+ 15
( 3 x 3+ 5 x 2 ) + ( 9 x +15 )
3) C. x 2 ( 3 x +5 ) +3 ( 3 x +5 )
4) D. ( x 2+ 3)(3 x +5)
18 x 4 +30 x3 + 42 x 2
6 x 2 ( 3 x 2 +5 x+7 )
5) A.
64 a3+ 27 b3
( 4 a )3 + ( 3 b )3 ADDITIONAL ACTIVITIES
( 4 a+3 b ) ( 16 a 2−12 ab+9 b2 )
Given:
6) A. V =351
3
216 x −125 l=4+ w
( 6 x )3−( 5 )3 w=w
(6 x−5)(36 x 2+ 30 x +25) 1
h= w
3
7) D.
5 x 3−45 x Solution:
5 x ( x 2−9 ) V =lwh
5 x [ ( x ) 2−( 3 )2 ]
5 x ( x−3 )( x +3 )
( 13 w )
351=( 4+ w ) ( w )
1
351=( 4+ w ) ( w ) 2
8) B. 3
x 2+ 40 x+ 400 4 1
351= w2 + w3
( x )2+ [ (2 )( x )( 20 ) ] + ( 20 )2 3 3
( x +20 ) ( x +20 )∨( x+ 20 )2 4 2 1 3
0= w + w −351
3 3
4 w + w3 −1053
9) A. 2
12 x2 +13 x−35 0=
3
12 x2 +28 x−15 x−35 0=4 w +w 3−1053
2
0=w 3+ 4 w 2−1053
Find h:
0=w 3−9 w 2+13 w2−1053 1
0=( w 3−9 w 2 ) + ( 13 w2 −1053 ) h= w
3
0=w 2 ( w−9 )+13 ( w2−81 ) 1
h= ( 9 )
0=w 2 ( w−9 )+13 ( w−9 )( w+9 ) 3
0=( w−9 ) [ w 2+13 ( w+9 ) ] 9
h=
0=( w−9 ) ( w 2+ 13 x +117 ) 3
h=3
Find w:
w−9=0 ∴ thedimensions of the cake pan should be
w=9 9 cm x 13 cm x 3 cm
Find l:
l=4+ w
l=4+ 9
l=13
Module 11
Illustrates Polynomial Equations
WHAT’S MORE
WHAT I HAVE LEARNED
WHAT I CAN DO
ASSESSMENT
ADDITIONAL ACTIVITIES
Module 12
Problems Involving Polynomials and Polynomial Equations
WHAT’S MORE
Activity 1. The Remains
Activity 2. Word Problem
WHAT I HAVE LEARNED
Activity. Do-Search-Learned
WHAT I CAN DO
Activity 1. Let’s Apply
Activity 2. Apply…Apply…
ASSESSMENT
ADDITIONAL ACTIVITIES
