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PROBLEMSET 
Submitted by : Fajardo, InnahGrace L. 
BSFT1-1D 
Submitted to: Prof. RegineCriseno 
Date: August 15, 2012
Factor the following. 
1.x5 –y5; x5+ y5 
2.x7–y7; x7+ y7 
3.x11–y11; x11+ y11 
4. Prove that 
xn–yn= (x –y) (... 
xn+ yn= (x –y) (...
1. x5 –y5; x5+ y5 
(x -y) (x4+ x3y + x2y2+ xy3+ y4) 
x5 –y5 
= 
x5+ y5 
= 
(x + y) (x4-x3y + x2y2-xy3+ y4) 
answers
2. x7–y7; x7+ y7 
answers 
x7–y7 
= 
(x –y) (x6+ x5y + x4y2+ x3y3+ x2y4+ xy5+ y6) 
x7+ y7 
= 
(x + y) (x6-x5y + x4y2-x3y3+ x2y4-xy5+ y6)
3. x11–y11; x11+ y11 
answersx11 –y11 
= 
(x –y) (x10+ x9y + x8y2+ x7y3+ x6y4+ x5y5+ x4y6+ x3y7 + x2y8+ xy9+y10) 
x11+ y11 
= 
(x + y) (x10-x9y + x8y2-x7y3+ x6y4-x5y5+ x4y6-x3y7 + x2y8-xy9+y10)
4. Prove that 
The first term would be (x -y) 
xn–yn= (x –y) (xn-1+ xn-2y +xn-3y2+ xn-4y3+ ... + yn-1) 
To prove that (x –y) is a factor of xn–yn, use factor theorem. 
Get the root of x -y which is y 
Use the factor theorem. 
P(x) = xn–yn 
P(y) = yn-yn 
P(y) = 0 
The answer is 0. This means that x –y is a factor of xn–yn. 
xn–yn= (x –y) (xn-1+ xn-2y +xn-3y2+ xn-4y3+ ... + yn-1)
The second factor is (xn-1+ xn-2y +xn-3y2+ xn-4y3+ ... + yn-1) 
In the second factor, the first term is xn-1. The exponent of the first term of the second factor is one less than the value of n. 
xn-1... 
The first power of y is 0 ( y0= 1) so y will only be seen on the second term. 
xn-1+ xn-2y ... 
The exponent of x decreases as the exponent of y increases. 
(xn-1+ xn-2y +xn-3y2+ xn-4y3+ ... + yn-1) 
The last term of the second factor of xn–ynwould be 
yn-1 
To determine its exponent, subtract 1 from the original value of n. 
The variable x is not present on the last term because its exponent there is 0 (x0= 1).
In the case of xn–yn, all the signs are + .(That insures the cancelling.) Example: x5 –y5= (x -y) (x4 + x3y + x2y2 + xy3 + y4) x5 –y5= x5+ x4y + x3y2 + x2y3+ xy4 –x4y –x3y2–x2y3–xy4–y5x5 –y5= x5 –y5 
xn+ yn= (x + y) (xn-1-xn-2y + xn-3y2-xn-4y3+ ... + yn-1) 
This is the same with the previous stated explanation. The only difference is the signs. The signs of the second factor alternates because if not, when we multiply the factors, nothing would be cancel. 
I used an example to prove it right. 
Example: 
x5+ y5 =(x + y) (x4-x3y + x2y2-xy3+ y4)
x5+ y5 = (x + y) (x4-x3y + x2y2-xy3+ y4) 
x5+ y5 = (x + y) (x4-x3y + x2y2-xy3+ y4) 
x5+ y5 = x5–x4y + x3y2–x2y3+ xy4+ x4y –x3y2 –x2y3–xy4+ y5 
x5+ y5 = x5+ y5

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Algebra [project]

  • 1. PROBLEMSET Submitted by : Fajardo, InnahGrace L. BSFT1-1D Submitted to: Prof. RegineCriseno Date: August 15, 2012
  • 2. Factor the following. 1.x5 –y5; x5+ y5 2.x7–y7; x7+ y7 3.x11–y11; x11+ y11 4. Prove that xn–yn= (x –y) (... xn+ yn= (x –y) (...
  • 3. 1. x5 –y5; x5+ y5 (x -y) (x4+ x3y + x2y2+ xy3+ y4) x5 –y5 = x5+ y5 = (x + y) (x4-x3y + x2y2-xy3+ y4) answers
  • 4. 2. x7–y7; x7+ y7 answers x7–y7 = (x –y) (x6+ x5y + x4y2+ x3y3+ x2y4+ xy5+ y6) x7+ y7 = (x + y) (x6-x5y + x4y2-x3y3+ x2y4-xy5+ y6)
  • 5. 3. x11–y11; x11+ y11 answersx11 –y11 = (x –y) (x10+ x9y + x8y2+ x7y3+ x6y4+ x5y5+ x4y6+ x3y7 + x2y8+ xy9+y10) x11+ y11 = (x + y) (x10-x9y + x8y2-x7y3+ x6y4-x5y5+ x4y6-x3y7 + x2y8-xy9+y10)
  • 6. 4. Prove that The first term would be (x -y) xn–yn= (x –y) (xn-1+ xn-2y +xn-3y2+ xn-4y3+ ... + yn-1) To prove that (x –y) is a factor of xn–yn, use factor theorem. Get the root of x -y which is y Use the factor theorem. P(x) = xn–yn P(y) = yn-yn P(y) = 0 The answer is 0. This means that x –y is a factor of xn–yn. xn–yn= (x –y) (xn-1+ xn-2y +xn-3y2+ xn-4y3+ ... + yn-1)
  • 7. The second factor is (xn-1+ xn-2y +xn-3y2+ xn-4y3+ ... + yn-1) In the second factor, the first term is xn-1. The exponent of the first term of the second factor is one less than the value of n. xn-1... The first power of y is 0 ( y0= 1) so y will only be seen on the second term. xn-1+ xn-2y ... The exponent of x decreases as the exponent of y increases. (xn-1+ xn-2y +xn-3y2+ xn-4y3+ ... + yn-1) The last term of the second factor of xn–ynwould be yn-1 To determine its exponent, subtract 1 from the original value of n. The variable x is not present on the last term because its exponent there is 0 (x0= 1).
  • 8. In the case of xn–yn, all the signs are + .(That insures the cancelling.) Example: x5 –y5= (x -y) (x4 + x3y + x2y2 + xy3 + y4) x5 –y5= x5+ x4y + x3y2 + x2y3+ xy4 –x4y –x3y2–x2y3–xy4–y5x5 –y5= x5 –y5 xn+ yn= (x + y) (xn-1-xn-2y + xn-3y2-xn-4y3+ ... + yn-1) This is the same with the previous stated explanation. The only difference is the signs. The signs of the second factor alternates because if not, when we multiply the factors, nothing would be cancel. I used an example to prove it right. Example: x5+ y5 =(x + y) (x4-x3y + x2y2-xy3+ y4)
  • 9. x5+ y5 = (x + y) (x4-x3y + x2y2-xy3+ y4) x5+ y5 = (x + y) (x4-x3y + x2y2-xy3+ y4) x5+ y5 = x5–x4y + x3y2–x2y3+ xy4+ x4y –x3y2 –x2y3–xy4+ y5 x5+ y5 = x5+ y5