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Factoring Perfect Square Trinomials Worksheet

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FACTORING TECHNIQUES WORKSHEET * Perfect Square Trinomials

1 of 2
Mathematics 8 Worksheet Algebra 1st
Quarter Mr. Carlo Justino J. Luna
Republic of the Philippines | DEPARTMENT OF EDUCATION
Region III | Division of City Schools | West District
Tamarind St., Clarkview Subd., Malabanias, Angeles City
School Year 2018-2019
NAME: ________________________________________ SCORE: __________________
GRADE & SECTION: _____________________________ DATE: ____________________
1st Quarter Worksheet #9
Recognizing the pattern to perfect squares isn't a make-or-break issue β€” these are quadratics that
you can factor in the usual way β€” but noticing the pattern can be a time-saver occasionally, which
can be helpful on timed tests.
The trick to seeing this pattern is really quite simple: If the first and third terms are squares, figure out
what they're squares of. Multiply those things, multiply that product by 2, and then compare your
result with the original quadratic's middle term. If you've got a match (ignoring the sign), then you've
got a perfect-square trinomial. And the original binomial that they'd squared was the sum (or
difference) of the square roots of the first and third terms, together with the sign that was on the
middle term of the trinomial. (http://www.purplemath.com/modules/specfact3.htm)
The patterns to remember when factoring perfect square trinomials are the following:
𝒂 𝟐
+ πŸπ’‚π’ƒ + 𝒃 𝟐
= ( 𝒂 + 𝒃) 𝟐
𝒂 𝟐
βˆ’ πŸπ’‚π’ƒ + 𝒃 𝟐
= (𝒂 βˆ’ 𝒃) 𝟐
Notice that all you have to do is to use the base of the first term and the last term. In the pattern just
described, the first term is π‘Ž2
and the base is π‘Ž, the last term is 𝑏2
and the base is 𝑏. Put the bases
inside parentheses with a 𝑝𝑙𝑒𝑠 between them (π‘Ž + 𝑏). Raise everything to the second power (π‘Ž + 𝑏)2
and you are done. Notice that I put a 𝑝𝑙𝑒𝑠 between π‘Ž and 𝑏. You will put a π‘šπ‘–π‘›π‘’π‘  if the second term is
π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’!
Factor each expression completely. (All are factorable.)
1. π‘Ž2
+ 12π‘Ž + 36 8. 4π‘Ž2
+ 20π‘Ž + 25
2. 𝑏2
+ 18𝑏 + 81 9. 9𝑏2
βˆ’ 60𝑏 + 100
3. 𝑐2
βˆ’ 22𝑐 + 121 10. 25𝑐2
+ 30𝑐 + 9
4. π‘š2
βˆ’ 24π‘š + 144 11. 16π‘š2
βˆ’ 40π‘š + 25
5. 𝑛2
+ 20𝑛 + 100 12. 36𝑛2
+ 132𝑛 + 121
6. π‘₯2
+ 10π‘₯ + 25 13. 49π‘₯2
βˆ’ 28π‘₯ + 4
7. 𝑦2
βˆ’ 8𝑦 + 16 14. 4𝑦2
+ 36𝑦 + 81
Mathematics 8 Worksheet Algebra 1st
Quarter Mr. Carlo Justino J. Luna
Republic of the Philippines | DEPARTMENT OF EDUCATION
Region III | Division of City Schools | West District
Tamarind St., Clarkview Subd., Malabanias, Angeles City
School Year 2018-2019
NAME: ________________________________________ SCORE: __________________
GRADE & SECTION: _____________________________ DATE: ____________________
1st Quarter Worksheet #9
Recognizing the pattern to perfect squares isn't a make-or-break issue β€” these are quadratics that
you can factor in the usual way β€” but noticing the pattern can be a time-saver occasionally, which
can be helpful on timed tests.
The trick to seeing this pattern is really quite simple: If the first and third terms are squares, figure out
what they're squares of. Multiply those things, multiply that product by 2, and then compare your
result with the original quadratic's middle term. If you've got a match (ignoring the sign), then you've
got a perfect-square trinomial. And the original binomial that they'd squared was the sum (or
difference) of the square roots of the first and third terms, together with the sign that was on the
middle term of the trinomial. (http://www.purplemath.com/modules/specfact3.htm)
The patterns to remember when factoring perfect square trinomials are the following:
𝒂 𝟐
+ πŸπ’‚π’ƒ + 𝒃 𝟐
= ( 𝒂 + 𝒃) 𝟐
𝒂 𝟐
βˆ’ πŸπ’‚π’ƒ + 𝒃 𝟐
= (𝒂 βˆ’ 𝒃) 𝟐
Notice that all you have to do is to use the base of the first term and the last term. In the pattern just
described, the first term is π‘Ž2
and the base is π‘Ž, the last term is 𝑏2
and the base is 𝑏. Put the bases
inside parentheses with a 𝑝𝑙𝑒𝑠 between them (π‘Ž + 𝑏). Raise everything to the second power (π‘Ž + 𝑏)2
and you are done. Notice that I put a 𝑝𝑙𝑒𝑠 between π‘Ž and 𝑏. You will put a π‘šπ‘–π‘›π‘’π‘  if the second term is
π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’!
Factor each expression completely. (All are factorable.)
1. π‘Ž2
+ 12π‘Ž + 36 8. 4π‘Ž2
+ 20π‘Ž + 25
= (π‘Ž + 6)2
= (2π‘Ž + 5)2
2. 𝑏2
+ 18𝑏 + 81 9. 9𝑏2
βˆ’ 60𝑏 + 100
= (𝑏 + 9)2
= (3𝑏 βˆ’ 10)2
3. 𝑐2
βˆ’ 22𝑐 + 121 10. 25𝑐2
+ 30𝑐 + 9
= (𝑐 βˆ’ 11)2
= (5𝑐 + 3)2
4. π‘š2
βˆ’ 24π‘š + 144 11. 16π‘š2
βˆ’ 40π‘š + 25
= (π‘š βˆ’ 12)2
= (4π‘š βˆ’ 5)2
5. 𝑛2
+ 20𝑛 + 100 12. 36𝑛2
+ 132𝑛 + 121
= (𝑛 + 10)2
= (6𝑛 + 11)2
6. π‘₯2
+ 10π‘₯ + 25 13. 49π‘₯2
βˆ’ 28π‘₯ + 4
= (π‘₯ + 5)2
= (7π‘₯ βˆ’ 2)2
7. 𝑦2
βˆ’ 8𝑦 + 16 14. 4𝑦2
+ 36𝑦 + 81
= (𝑦 βˆ’ 4)2
= (2𝑦 + 9)2

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  • 1. Mathematics 8 Worksheet Algebra 1st Quarter Mr. Carlo Justino J. Luna Republic of the Philippines | DEPARTMENT OF EDUCATION Region III | Division of City Schools | West District Tamarind St., Clarkview Subd., Malabanias, Angeles City School Year 2018-2019 NAME: ________________________________________ SCORE: __________________ GRADE & SECTION: _____________________________ DATE: ____________________ 1st Quarter Worksheet #9 Recognizing the pattern to perfect squares isn't a make-or-break issue β€” these are quadratics that you can factor in the usual way β€” but noticing the pattern can be a time-saver occasionally, which can be helpful on timed tests. The trick to seeing this pattern is really quite simple: If the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2, and then compare your result with the original quadratic's middle term. If you've got a match (ignoring the sign), then you've got a perfect-square trinomial. And the original binomial that they'd squared was the sum (or difference) of the square roots of the first and third terms, together with the sign that was on the middle term of the trinomial. (http://www.purplemath.com/modules/specfact3.htm) The patterns to remember when factoring perfect square trinomials are the following: 𝒂 𝟐 + πŸπ’‚π’ƒ + 𝒃 𝟐 = ( 𝒂 + 𝒃) 𝟐 𝒂 𝟐 βˆ’ πŸπ’‚π’ƒ + 𝒃 𝟐 = (𝒂 βˆ’ 𝒃) 𝟐 Notice that all you have to do is to use the base of the first term and the last term. In the pattern just described, the first term is π‘Ž2 and the base is π‘Ž, the last term is 𝑏2 and the base is 𝑏. Put the bases inside parentheses with a 𝑝𝑙𝑒𝑠 between them (π‘Ž + 𝑏). Raise everything to the second power (π‘Ž + 𝑏)2 and you are done. Notice that I put a 𝑝𝑙𝑒𝑠 between π‘Ž and 𝑏. You will put a π‘šπ‘–π‘›π‘’π‘  if the second term is π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’! Factor each expression completely. (All are factorable.) 1. π‘Ž2 + 12π‘Ž + 36 8. 4π‘Ž2 + 20π‘Ž + 25 2. 𝑏2 + 18𝑏 + 81 9. 9𝑏2 βˆ’ 60𝑏 + 100 3. 𝑐2 βˆ’ 22𝑐 + 121 10. 25𝑐2 + 30𝑐 + 9 4. π‘š2 βˆ’ 24π‘š + 144 11. 16π‘š2 βˆ’ 40π‘š + 25 5. 𝑛2 + 20𝑛 + 100 12. 36𝑛2 + 132𝑛 + 121 6. π‘₯2 + 10π‘₯ + 25 13. 49π‘₯2 βˆ’ 28π‘₯ + 4 7. 𝑦2 βˆ’ 8𝑦 + 16 14. 4𝑦2 + 36𝑦 + 81
  • 2. Mathematics 8 Worksheet Algebra 1st Quarter Mr. Carlo Justino J. Luna Republic of the Philippines | DEPARTMENT OF EDUCATION Region III | Division of City Schools | West District Tamarind St., Clarkview Subd., Malabanias, Angeles City School Year 2018-2019 NAME: ________________________________________ SCORE: __________________ GRADE & SECTION: _____________________________ DATE: ____________________ 1st Quarter Worksheet #9 Recognizing the pattern to perfect squares isn't a make-or-break issue β€” these are quadratics that you can factor in the usual way β€” but noticing the pattern can be a time-saver occasionally, which can be helpful on timed tests. The trick to seeing this pattern is really quite simple: If the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2, and then compare your result with the original quadratic's middle term. If you've got a match (ignoring the sign), then you've got a perfect-square trinomial. And the original binomial that they'd squared was the sum (or difference) of the square roots of the first and third terms, together with the sign that was on the middle term of the trinomial. (http://www.purplemath.com/modules/specfact3.htm) The patterns to remember when factoring perfect square trinomials are the following: 𝒂 𝟐 + πŸπ’‚π’ƒ + 𝒃 𝟐 = ( 𝒂 + 𝒃) 𝟐 𝒂 𝟐 βˆ’ πŸπ’‚π’ƒ + 𝒃 𝟐 = (𝒂 βˆ’ 𝒃) 𝟐 Notice that all you have to do is to use the base of the first term and the last term. In the pattern just described, the first term is π‘Ž2 and the base is π‘Ž, the last term is 𝑏2 and the base is 𝑏. Put the bases inside parentheses with a 𝑝𝑙𝑒𝑠 between them (π‘Ž + 𝑏). Raise everything to the second power (π‘Ž + 𝑏)2 and you are done. Notice that I put a 𝑝𝑙𝑒𝑠 between π‘Ž and 𝑏. You will put a π‘šπ‘–π‘›π‘’π‘  if the second term is π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’! Factor each expression completely. (All are factorable.) 1. π‘Ž2 + 12π‘Ž + 36 8. 4π‘Ž2 + 20π‘Ž + 25 = (π‘Ž + 6)2 = (2π‘Ž + 5)2 2. 𝑏2 + 18𝑏 + 81 9. 9𝑏2 βˆ’ 60𝑏 + 100 = (𝑏 + 9)2 = (3𝑏 βˆ’ 10)2 3. 𝑐2 βˆ’ 22𝑐 + 121 10. 25𝑐2 + 30𝑐 + 9 = (𝑐 βˆ’ 11)2 = (5𝑐 + 3)2 4. π‘š2 βˆ’ 24π‘š + 144 11. 16π‘š2 βˆ’ 40π‘š + 25 = (π‘š βˆ’ 12)2 = (4π‘š βˆ’ 5)2 5. 𝑛2 + 20𝑛 + 100 12. 36𝑛2 + 132𝑛 + 121 = (𝑛 + 10)2 = (6𝑛 + 11)2 6. π‘₯2 + 10π‘₯ + 25 13. 49π‘₯2 βˆ’ 28π‘₯ + 4 = (π‘₯ + 5)2 = (7π‘₯ βˆ’ 2)2 7. 𝑦2 βˆ’ 8𝑦 + 16 14. 4𝑦2 + 36𝑦 + 81 = (𝑦 βˆ’ 4)2 = (2𝑦 + 9)2