Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods.
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions.
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (Β± ) + 3*(Β± ) = 5.
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (Β± ) + 3*(Β± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (Β± ) + 3*(Β± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (Β± ) + 3* (Β± ) = β5.
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (Β± ) + 3*(Β± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (Β± ) + 3* (Β± ) = β5.
Yes, 1* (1) + 3* (β2) = β5
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (Β± ) + 3*(Β± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (Β± ) + 3* (Β± ) = β5.
Yes, 1* (1) + 3* (β2) = β5 or 1* (β2) + 3* (β1) = β5
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (Β± ) + 3*(Β± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (Β± ) + 3* (Β± ) = β5.
Yes, 1* (1) + 3* (β2) = β5 or 1* (β2) + 3* (β1) = β5
c. 1* (Β± ) + 3* (Β± ) = 8.
Factoring Trinomials II
Now letβs try to factor trinomials of the form ax2 + bx + c.
Weβll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (Β± ) + 3*(Β± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (Β± ) + 3* (Β± ) = β5.
Yes, 1* (1) + 3* (β2) = β5 or 1* (β2) + 3* (β1) = β5
c. 1* (Β± ) + 3* (Β± ) = 8.
No, since the most we can obtain is 1* (1) + 3* (2) = 7.
Factoring Trinomials II
(Reversed FOIL Method)
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2 to get the constant term +2.
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x Β± #)(1x Β± #) must yields +5x,
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x Β± #)(1x Β± #) must yields +5x, or that
3(Β± # ) +1(Β± #) = 5 where the #βs are 1 and 2.
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x Β± #)(1x Β± #) must yields +5x, or that
3(Β± # ) +1(Β± #) = 5 where the #βs are 1 and 2.
Since 3(1) +1(2) = 5,
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x Β± #)(1x Β± #) must yields +5x, or that
3(Β± # ) +1(Β± #) = 5 where the #βs are 1 and 2.
Since 3(1) +1(2) = 5,
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x Β± #)(1x Β± #) must yields +5x, or that
3(Β± # ) +1(Β± #) = 5 where the #βs are 1 and 2.
Since 3(1) +1(2) = 5,
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x Β± #)(1x Β± #) must yields +5x, or that
3(Β± # ) +1(Β± #) = 5 where the #βs are 1 and 2.
Since 3(1) +1(2) = 5, we see that
3x2 + 5x + 2 = (3x + 2)(1x + 1).
Factoring Trinomials II
(Reversed FOIL Method)
Letβs see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x Β± #)(1x Β± #).
The #βs must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x Β± #)(1x Β± #) must yields +5x, or that
3(Β± # ) +1(Β± #) = 5 where the #βs are 1 and 2.
Since 3(1) +1(2) = 5, we see that
3x2 + 5x + 2 = (3x + 2)(1x + 1).
5x
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
It's 3(β2) + 1(β1) = β7.
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
It's 3(β2) + 1(β1) = β7.
So 3x2 β 7x + 2 = (3x β1)(1x β 2)
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
It's 3(β2) + 1(β1) = β7.
So 3x2 β 7x + 2 = (3x β1)(1x β 2)
Example D. Factor 3x2 + 5x β 2.
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
It's 3(β2) + 1(β1) = β7.
So 3x2 β 7x + 2 = (3x β1)(1x β 2)
Example D. Factor 3x2 + 5x β 2.
We start with (3x Β± #)(1x Β± #).
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
It's 3(β2) + 1(β1) = β7.
So 3x2 β 7x + 2 = (3x β1)(1x β 2)
Example D. Factor 3x2 + 5x β 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
It's 3(β2) + 1(β1) = β7.
So 3x2 β 7x + 2 = (3x β1)(1x β 2)
Example D. Factor 3x2 + 5x β 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = +5.
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
It's 3(β2) + 1(β1) = β7.
So 3x2 β 7x + 2 = (3x β1)(1x β 2)
Example D. Factor 3x2 + 5x β 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = +5.
Since c is negative, they must have opposite signs .
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
It's 3(β2) + 1(β1) = β7.
So 3x2 β 7x + 2 = (3x β1)(1x β 2)
Example D. Factor 3x2 + 5x β 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = +5.
Since c is negative, they must have opposite signs .
It is 3(+2) + 1(β1) = +5.
Factoring Trinomials II
Example C. Factor 3x2 β 7x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = β7.
It's 3(β2) + 1(β1) = β7.
So 3x2 β 7x + 2 = (3x β1)(1x β 2)
Example D. Factor 3x2 + 5x β 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1 and 2 as #'s so that
3(Β± # ) + 1(Β± # ) = +5.
Since c is negative, they must have opposite signs .
It is 3(+2) + 1(β1) = +5.
So 3x2 + 5x + 2 = (3x β1)(1x + 2)
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible.
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x β 4.
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x β 4.
We start with (3x Β± #)(1x Β± #).
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x β 4.
We start with (3x Β± #)(1x Β± #). Since 4 = 2(2) = 1(4),
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x β 4.
We start with (3x Β± #)(1x Β± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that
3(Β± # ) + 1(Β± # ) = +11.
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x β 4.
We start with (3x Β± #)(1x Β± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that
3(Β± # ) + 1(Β± # ) = +11.
It can't be 2&2.
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x β 4.
We start with (3x Β± #)(1x Β± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that
3(Β± # ) + 1(Β± # ) = +11.
It can't be 2&2.
Try 1&4,
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x β 4.
We start with (3x Β± #)(1x Β± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that
3(Β± # ) + 1(Β± # ) = +11.
It can't be 2&2.
Try 1&4, it is
3(+4) + 1(β1) = +11.
Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2.
We start with (3x Β± #)(1x Β± #).
We need to fill in 1&2 so that
3(Β± # ) + 1(Β± # ) = +8.
This is impossible. Hence the expression is prime.
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x β 4.
We start with (3x Β± #)(1x Β± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that
3(Β± # ) + 1(Β± # ) = +11. It can't be 2&2.
Try 1&4, it is
3(+4) + 1(β1) = +11.
So 3x2 + 11x β 4 = (3x β 1)(1x + 4).
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Since 3 must be 3(1),
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
(Β± #)(Β± 3) + (Β± #)(Β±1) = β 5.
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
(Β± #)(Β± 3) + (Β± #)(Β±1) = β 5.
12 = 1(12) = 2(6) = 3(4)
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
(Β± #)(Β± 3) + (Β± #)(Β±1) = β 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
(Β± #)(Β± 3) + (Β± #)(Β±1) = β 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
We get (3)(β3) + (4)(+1) = β 5.
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
(Β± #)(Β± 3) + (Β± #)(Β±1) = β 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
We get (3)(β3) + (4)(+1) = β 5.
So 12x2 β 5x β 3 = (3x + 1)(4x β 3).
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
(Β± #)(Β± 3) + (Β± #)(Β±1) = β 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
We get (3)(β3) + (4)(+1) = β 5.
So 12x2 β 5x β 3 = (3x + 1)(4x β 3).
Remark:
In the above method, finding
(#)(Β± #) + (#)( Β± #) = b
does not guarantee that the trinomial will factor.
Factoring Trinomials II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
Example G. Factor 12x2 β 5x β 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
(Β± #)(Β± 3) + (Β± #)(Β±1) = β 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
We get (3)(β3) + (4)(+1) = β 5.
So 12x2 β 5x β 3 = (3x + 1)(4x β 3).
Remark:
In the above method, finding
(#)(Β± #) + (#)( Β± #) = b
does not guarantee that the trinomial will factor. We have to
match the sign of c also.
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
We start with (3x Β± #)(1x Β± #).
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
We start with (3x Β± #)(1x Β± #). We find that:
3(β2) + 1(β1) = β7.
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
We start with (3x Β± #)(1x Β± #). We find that:
3(β2) + 1(β1) = β7.
But this won't work since (β2)(β1) = 2 = c.
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
We start with (3x Β± #)(1x Β± #). We find that:
3(β2) + 1(β1) = β7.
But this won't work since (β2)(β1) = 2 = c.
In fact this trinomial is prime.
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
We start with (3x Β± #)(1x Β± #). We find that:
3(β2) + 1(β1) = β7.
But this won't work since (β2)(β1) = 2 = c.
In fact this trinomial is prime.
There might be multiple matchings for
(#)(Β± #) + (#)( Β± #) = b
make sure you chose the correct one, if any.
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
We start with (3x Β± #)(1x Β± #). We find that:
3(β2) + 1(β1) = β7.
But this won't work since (β2)(β1) = 2 = c.
In fact this trinomial is prime.
There might be multiple matchings for
(#)(Β± #) + (#)( Β± #) = b
make sure you chose the correct one, if any.
Example I: Factor 1x2 + 5x β 6 .
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
We start with (3x Β± #)(1x Β± #). We find that:
3(β2) + 1(β1) = β7.
But this won't work since (β2)(β1) = 2 = c.
In fact this trinomial is prime.
There might be multiple matchings for
(#)(Β± #) + (#)( Β± #) = b
make sure you chose the correct one, if any.
Example I: Factor 1x2 + 5x β 6 .
We have:
1(+3) + 1(+2) = +5
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
We start with (3x Β± #)(1x Β± #). We find that:
3(β2) + 1(β1) = β7.
But this won't work since (β2)(β1) = 2 = c.
In fact this trinomial is prime.
There might be multiple matchings for
(#)(Β± #) + (#)( Β± #) = b
make sure you chose the correct one, if any.
Example I: Factor 1x2 + 5x β 6 .
We have:
1(+3) + 1(+2) = +5
1(+6) + 1(β1) = +5
Factoring Trinomials II
Example H. Factor 3x2 β 7x β 2 .
We start with (3x Β± #)(1x Β± #). We find that:
3(β2) + 1(β1) = β7.
But this won't work since (β2)(β1) = 2 = c.
In fact this trinomial is prime.
There might be multiple matchings for
(#)(Β± #) + (#)( Β± #) = b
make sure you chose the correct one, if any.
Example I: Factor 1x2 + 5x β 6 .
We have:
1(+3) + 1(+2) = +5
1(+6) + 1(β1) = +5
The one that works is x2 + 5x β 6 = (x + 6)(x β 1).
Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure
Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor βx3 + 3x + 2x2
Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor βx3 + 3x + 2x2
βx3 + 3x + 2x2 Arrange the terms in order
= βx3 + 2x2 + 3x
Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor βx3 + 3x + 2x2
βx3 + 3x + 2x2 Arrange the terms in order
= βx3 + 2x2 + 3x Factor out the GCF
= β x(x2 β 2x β 3)
Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor βx3 + 3x + 2x2
βx3 + 3x + 2x2 Arrange the terms in order
= βx3 + 2x2 + 3x Factor out the GCF
= β x(x2 β 2x β 3)
= β x(x β 3)(x + 1)
Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor βx3 + 3x + 2x2
βx3 + 3x + 2x2 Arrange the terms in order
= βx3 + 2x2 + 3x Factor out the GCF
= β x(x2 β 2x β 3)
= β x(x β 3)(x + 1)
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