50 solving equations by factoring

Solve Equations by Factoring

Example A.We ordered pizzas to be delivered from Pizza Grande. Each pizza is $3 and there is $10 delivery charge. Suppose we paid $34 to the delivery guy at the door, how many pizzas should be there?


Example A.We ordered pizzas to be delivered from Pizza Grande. Each pizza is $3 and there is $10 delivery charge. Suppose we paid $34 to the delivery guy at the door, how many pizzas should be there?We backtrack the calculation by subtracting the $10 for the delivery to conclude that the pizzas cost $24.


Example A.We ordered pizzas to be delivered from Pizza Grande. Each pizza is $3 and there is $10 delivery charge. Suppose we paid $34 to the delivery guy at the door, how many pizzas should be there?We backtrack the calculation by subtracting the $10 for the delivery to conclude that the pizzas cost $24. Since each pizza is $3, there should be 8 pizzas.




Using symbols, let x represents the number of pizzas delivered.



Using symbols, let x represents the number of pizzas delivered. Since each pizza cost $3, so it cost 3x for x pizzas, add on the delivery charge, we have 3x + 10 as the total cost.


We paid $34 in total hence 3x + 10 = 34




We paid $34 in total hence 3x + 10 = 34 Subtract 10 –10 –10 3x = 24




We paid $34 in total hence 3x + 10 = 34 Subtract 10 –10 –10 3x = 24 Divide by 3 x = 8or that there should be 8 pizzas delivered.



The mathematical formulation “3x + 10 = 34” in an equation. Solve Equations by Factoring

The mathematical formulation “3x + 10 = 34” in an equation. An equations is a mathematical declaration of the form: “something = something (else)” or “left expression = right expression”


The mathematical formulation “3x + 10 = 34” in an equation. An equations is a mathematical declaration of the form: “something = something (else)” or “left expression = right expression”The value x = 8 is a solution for the equation “3x + 10 = 34” since 3(8) + 10 is (=) 34.


The mathematical formulation “3x + 10 = 34” in an equation. An equations is a mathematical declaration of the form: “something = something (else)” or “left expression = right expression”The value x = 8 is a solution for the equation “3x + 10 = 34” since 3(8) + 10 is (=) 34. It’s the only solution of “3x + 10 = 34”.


The mathematical formulation “3x + 10 = 34” in an equation. An equations is a mathematical declaration of the form: “something = something (else)” or “left expression = right expression”The value x = 8 is a solution for the equation “3x + 10 = 34” since 3(8) + 10 is (=) 34. It’s the only solution of “3x + 10 = 34”.Where as expressions such as “3x + 10” are used to calculate outcomes (the cost for x pizzas delivered),


The mathematical formulation “3x + 10 = 34” in an equation. An equations is a mathematical declaration of the form: “something = something (else)” or “left expression = right expression”The value x = 8 is a solution for the equation “3x + 10 = 34” since 3(8) + 10 is (=) 34. It’s the only solution of “3x + 10 = 34”.Where as expressions such as “3x + 10” are used to calculate outcomes (the cost for x pizzas delivered),equations such as “3x + 10 = 34” are formulated from known outcomes for the purpose of backtracking to the original input x.


The mathematical formulation “3x + 10 = 34” in an equation. An equations is a mathematical declaration of the form: “something = something (else)” or “left expression = right expression”The value x = 8 is a solution for the equation “3x + 10 = 34” since 3(8) + 10 is (=) 34. It’s the only solution of “3x + 10 = 34”.Where as expressions such as “3x + 10” are used to calculate outcomes (the cost for x pizzas delivered),equations such as “3x + 10 = 34” are formulated from known outcomes for the purpose of backtracking to the original input x. “3x + 10 = 34” is a first degree or linear equation because the highest degree of x appeared in the equation is 1.




The Solution of Linear EquationsThe solution for ax + b = 0 (so that ax = –b) is x = –b/a (a≠0).



The Solution of Linear EquationsThe solution for ax + b = 0 (so that ax = –b) is x = –b/a (a≠0). Example B. The solution fora. 3x + 10 = 0b. 2x – 10 = 0



The Solution of Linear EquationsThe solution for ax + b = 0 (so that ax = –b) is x = –b/a (a≠0). Example B. The solution fora. 3x + 10 = 0 is x = –10/3 b. 2x – 10 = 0



The Solution of Linear EquationsThe solution for ax + b = 0 (so that ax = –b) is x = –b/a (a≠0). Example B. The solution fora. 3x + 10 = 0 is x = –10/3 b. 2x – 10 = 0 is x = – (–10)/2 = 5

Solve Equations by FactoringPolynomial equations are equations of the form

polynomial = polynomial


polynomial = polynomial An example of a polynomial equation is

x2 = 1



Given an polynomial equation, after collecting all the terms to one side, the equation may be simplified as “# = 0” or as “0 = #”, which are the 0–forms of the equation.

An example of a polynomial equation isx2 = 1



Given an polynomial equation, after collecting all the terms to one side, the equation may be simplified as “# = 0” or as “0 = #”, which are the 0–forms of the equation. For example, the 0–forms of the equation x2 = 1 are

x2 –1 = 0 and 0 = 1 – x2





x2 –1 = 0 and 0 = 1 – x2


The highest degree of x in the 0–forms of an equation is the degree of the equation.




x2 –1 = 0 and 0 = 1 – x2


The highest degree of x in the 0–forms of an equation is the degree of the equation. Hence x2 –1 = 0 is a 2nd degree equation




x2 –1 = 0 and 0 = 1 – x2


The highest degree of x in the 0–forms of an equation is the degree of the equation. Hence x2 –1 = 0 is a 2nd degree equation and that 2x4 + 1 = 2x4 + x3 is a 3rd degree equation.




x2 –1 = 0 and 0 = 1 – x2


The highest degree of x in the 0–forms of an equation is the degree of the equation. Hence x2 –1 = 0 is a 2nd degree equation and that 2x4 + 1 = 2x4 + x3 is a 3rd degree equation.

The significance of the 0–form is based on the “property of 0” – that when multiplying numbers, the only way to obtain 0 as their product is that one (or more) of the numbers is 0.

Solve Equations by FactoringThe 0–Product PropertyIf AB = 0 or 0 = AB, then either A = 0 or B = 0.

Solve Equations by FactoringThe 0–Product PropertyIf AB = 0 or 0 = AB, then either A = 0 or B = 0. In general, if A*B*C* .. = 0 then either A = 0 or B = 0 or C = 0,..


Example C. a. If 3x = 0, then x = 0 is the only solution because 3 ≠ 0.

The 0–Product PropertyIf AB = 0 or 0 = AB, then either A = 0 or B = 0. In general, if A*B*C* .. = 0 then either A = 0 or B = 0 or C = 0,..


Example C. a. If 3x = 0, then x = 0 is the only solution because 3 ≠ 0. b. If 3(x – 2) = 0,



Example C. a. If 3x = 0, then x = 0 is the only solution because 3 ≠ 0. b. If 3(x – 2) = 0, because 3 ≠ 0, so (x – 2) = 0



Example C. a. If 3x = 0, then x = 0 is the only solution because 3 ≠ 0. b. If 3(x – 2) = 0, because 3 ≠ 0, so (x – 2) = 0 or that x = 2 is the only solution



Example C. a. If 3x = 0, then x = 0 is the only solution because 3 ≠ 0. b. If 3(x – 2) = 0, because 3 ≠ 0, so (x – 2) = 0 or that x = 2 is the only solution c. If 3x(x – 2) = 0, because 3 ≠ 0, then either x = 0,



Example C. a. If 3x = 0, then x = 0 is the only solution because 3 ≠ 0. b. If 3(x – 2) = 0, because 3 ≠ 0, so (x – 2) = 0 or that x = 2 is the only solution c. If 3x(x – 2) = 0, because 3 ≠ 0, then either x = 0, or (x – 2) = 0 so x = 2.



Example C. a. If 3x = 0, then x = 0 is the only solution because 3 ≠ 0. b. If 3(x – 2) = 0, because 3 ≠ 0, so (x – 2) = 0 or that x = 2 is the only solution c. If 3x(x – 2) = 0, because 3 ≠ 0, then either x = 0, or (x – 2) = 0 so x = 2. There are two solutions {0,2} because there are two linear factors in x.





d. If 0 = x(3x + 10)(2x –10), then either x = 0




d. If 0 = x(3x + 10)(2x –10), then either x = 0 or 3x +10 = 0 so x = –10/3




d. If 0 = x(3x + 10)(2x –10), then either x = 0 or 3x +10 = 0 or 2x – 10 = 0 so x = –10/3 so x = 10/2 = 5




d. If 0 = x(3x + 10)(2x –10), then either x = 0 or 3x +10 = 0 or 2x – 10 = 0 so x = –10/3 so x = 10/2 = 5 There are three solutions {0,–10/3,5} because there are three x–linear factors.

Solve Equations by FactoringSolve Equations by Factoring

Solve Equations by FactoringSolve Equations by Factoring

Example D. Solve the equation for x. a. 3 = x2 – 2x


In view of example C, to solve an equation by factoring: Solve Equations by Factoring



In view of example C, to solve an equation by factoring: 1. convert the equation to the 0–form (set one side of the equation to 0 by moving all the terms to the other side.)*







* When setting one side of the equation to be 0, keep the highest degree x–term positive in preparation for the factoring.





Set the left side to be 0 (to keep the x2 positive for factoring) 0 = x2 – 2x – 3



In view of example C, to solve an equation by factoring: 1. convert the equation to the 0–form (set one side of the equation to 0 by moving all the terms to the other side.)* 2. factor the polynomial, if possible,



Set the left side to be 0 (to keep the x2 positive for factoring) 0 = x2 – 2x – 3



In view of example C, to solve an equation by factoring: 1. convert the equation to the 0–form (set one side of the equation to 0 by moving all the terms to the other side.)* 2. factor the polynomial, if possible,



Set the left side to be 0 (to keep the x2 positive for factoring) 0 = x2 – 2x – 3 Factor 0 = (x – 3)(x + 1)



In view of example C, to solve an equation by factoring: 1. convert the equation to the 0–form (set one side of the equation to 0 by moving all the terms to the other side.)* 2. factor the polynomial, if possible, 3. extract an answer from each linear factor.



Set the left side to be 0 (to keep the x2 positive for factoring) 0 = x2 – 2x – 3 Factor 0 = (x – 3)(x + 1)


There are two linear x–factors so we may extract an answer from each.





Set the left side to be 0 (to keep the x2 positive for factoring) 0 = x2 – 2x – 3 Factor 0 = (x – 3)(x + 1)Hence x – 3 = 0 or x + 1 = 0







Set the left side to be 0 (to keep the x2 positive for factoring) 0 = x2 – 2x – 3 Factor 0 = (x – 3)(x + 1)Hence x – 3 = 0 or x + 1 = 0 x = 3 or x = – 1



b. 2x(x + 1) = 4x + 3(1 – x) Solve Equations by Factoring

b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x


b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3


b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0


b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0


b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0


b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 or x – 1 = 0


b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = –3

or x – 1 = 0


b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = –3 x = –3/2

or x – 1 = 0



or x – 1 = 0 x = 1



or x – 1 = 0 x = 1

c. 8x(x2 – 1) = 10x



or x – 1 = 0 x = 1

c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x



or x – 1 = 0 x = 1

c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0



or x – 1 = 0 x = 1

c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0



or x – 1 = 0 x = 1

c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor

2x(2x + 3)(2x – 3) = 0



or x – 1 = 0 x = 1


2x(2x + 3)(2x – 3) = 0

There are three linear x–factors. We may extract one answer from each.



or x – 1 = 0 x = 1


2x(2x + 3)(2x – 3) = 0 x = 0




or x – 1 = 0 x = 1


2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0




or x – 1 = 0 x = 1


2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0




or x – 1 = 0 x = 1


2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0 2x = –3




or x – 1 = 0 x = 1


2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0 2x = –3 x = –3/2




or x – 1 = 0 x = 1


2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0 2x = –3 2x = 3 x = –3/2




or x – 1 = 0 x = 1


2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0 2x = –3 2x = 3 x = –3/2 x = 3/2



From the above discussions we see that the problem of finding solutions of a polynomial equations is the same problem as factoring the polynomials into linear factors.




However, most polynomials are not factorable by our methods.



However, most polynomials are not factorable by our methods. To utilize the ac–method, the coefficients of the 2nd degree equation have to be integers.



However, most polynomials are not factorable by our methods. To utilize the ac–method, the coefficients of the 2nd degree equation have to be integers. The ac–method only able to solve 2nd degree equations where the solution that are fractions.



However, most polynomials are not factorable by our methods. To utilize the ac–method, the coefficients of the 2nd degree equation have to be integers. The ac–method only able to solve 2nd degree equations where the solution that are fractions.On the hands, we may solve 2nd degree or quadratic equations by the the quadratic formula.



However, most polynomials are not factorable by our methods. To utilize the ac–method, the coefficients of the 2nd degree equation have to be integers. The ac–method only able to solve 2nd degree equations where the solution that are fractions.On the hands, we may solve 2nd degree or quadratic equations by the the quadratic formula. The Solution of Quadratic (2nd degree) Equations– The Quadratic Formula



However, most polynomials are not factorable by our methods. To utilize the ac–method, the coefficients of the 2nd degree equation have to be integers. The ac–method only able to solve 2nd degree equations where the solution that are fractions.On the hands, we may solve 2nd degree or quadratic equations by the the quadratic formula. The Solution of Quadratic (2nd degree) Equations– The Quadratic Formula

–b ± b2 – 4ac2ax =

The solution for ax2 + bx + c = 0 are

Solve Equations by FactoringExample F. Solve 3 = x2 – 2x by the Quadratic Formula.


Set the left side to be 0 so that 0 = x2 – 2x – 3,


Set the left side to be 0 so that 0 = x2 – 2x – 3, we have that a = 1, b = –2, c = –3so that (–2)2 – 4(1)(–3) = 16.


Set the left side to be 0 so that 0 = x2 – 2x – 3,we have that a = 1, b = –2, c = –3so that (–2)2 – 4(1)(–3) = 16. By the quadratic formula

–(–2) + 162(1)x = –(–2) – 16

2(1)x =,



–(–2) + 162(1)x = –(–2) – 16

2(1)x =,

2 + 42x =or ,



–(–2) + 162(1)x = –(–2) – 16

2(1)x =,

So x = 3

2 + 42x =or ,



–(–2) + 162(1)x = –(–2) – 16

2(1)x =,

So x = 3 or x = – 1

2 + 42x = x =,

2 – 42or

Ex. Solve the following equations. Check your answers.

19. x2 – 3x = 10 20. x(x – 2) = 24 21. 2x2 = 3(x + 1) – 1

28. x3 – 2x2 = 0

22. x2 = 425. 2x(x – 3) + 4 = 2x – 4

29. x3 – 2x2 – 8x = 031. 4x2 = x3

30. 2x2(x – 3) = –4x

26. x(x – 3) + x + 6 = 2x2 + 3x

13. x2 – 3x – 4 = 0 14. x2 – 2x – 15 = 0 15. x2 + 7x + 12 = 016. –x2 – 2x + 8 = 0 17. 9 – x2 = 0 18. 2x2 – x – 1 = 0

27. x(x + 4) + 9 = 2(2 – x)

23. 8x2 = 2 24. 27x2 – 12 = 0

32. 4x = x3 33. 4x2 = x4

34. 7x2 = –4x3 – 3x 35. 5 = (x + 2)(2x + 1)36. (x – 1)2 = (x + 1)2 – 4 37. (x + 1)2 = x2 + (x – 1)2

38. (x + 2)2 – (x + 1)2= x2 39. (x + 3)2 – (x + 2)2 = (x + 1)2


50 solving equations by factoring

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