Factoring Using the AC Method

10G (2.6) PART 2 Factoring using the AC method

Multiple Choice

Before, we had a quadratic equation that looked like this:

$$x^2+2x-15=0$$

How did we find the values of x?

• We rewrote it as

  $$x^{2\ }+5x-3x-15$$

  and

  factored it into

  (x + 5)(x - 3)

• We added 15 to both sides

  then we subtracted 2x from both sides

  then we took the root of both sides

Multiple Choice

Can we solve this quadratic equation in the same way?

$$3x^2+2x-15=0$$

• Yes but we need to use the AC Method.

• No because we can't factor it any further than

  $$3x\left(x-1\right)+5\left(x-3\right)$$

You need to use the AC Method to solve!
You've actually be using it already

​the a

​the c

Take note of the steps,





the ac method will make more sense when we do an example

Multiple Choice

Standard form of a polynomial is $$ax^2+bx+c=0$$

Let's try to do an example step by step:

$$2x^{2\ }-11x+5$$

STEP1: Identify a, b, and c.

• a = 2

  b = -11

  c = 5

• a = 3

  b = -11

  c = 5

• a = 2

  b = 11

  c = 1

• a = 2

  b = 1

  c = 5

• a = 1

  b = -11

  c = 5

Multiple Choice

$$2x^{2\ }-11x+5$$

a = 2

b = -11

c = 5

STEP 2: Multiply a and c to get our k

• 2(5) = 10

  so k = 10

• 2(-11) = -22

  so k = -22

• -11(5) = -55

  so k = -55

Multiple Select

$$2x^{2\ }-11x+5$$

k = 2(5) = 10

STEP 3: Find all the factor pairs of k

• 1(10)

  and

  -1(-10)

• 2(5)

  and

  -2(-5)

• 3(3)

  and

  -3(-3)

• 2(8)

  ana

  -2(-8)

Multiple Choice

$$2x^{2\ }-11x+5$$

the factor pairs of k are

1(10) and 2(5) and -2(-5) and -1(-10)

STEP 4: which one of these factor pairs add up to b = -11

• -1(-10)

  because

  -1 - 10 = -11

• 1(10)

  because

  1 + 10 = -11

• -2(5)

  because

  -2 - 5 = -11

• 2(5)

  because

  2 + 5 = -11

Multiple Choice

$$2x^{2\ }-11x+5$$

the factor pair we picked was -1(-10) because

-1 - 10 = -11, which is our b constant.

STEP 5: rewrite our quadratic expression using -10 and -1 instead of -11

• $$2x^{2\ }-10x\ -\ 1x+5$$

• $$2x^{2\ }+10x+1x+5$$

• $$2x^{2\ }-10x\ +1x+5$$

• $$2x^{2\ }+10x-1x+5$$

Multiple Choice

$$2x^{2\ }-11x+5$$

We rewrote the quadratic equation as

$$2x^{2\ }-10x-1x+5$$

STEP 6: separate our rewritten quadratic equation using two sets of parentheses.

e.g. (something) + (something)

• $$\left(2x^{2\ }-10x\right)\ +\left(-1x+5\right)$$

• $$\left(2x^{2\ }+\left(10x+1x\right)+5\right)$$

Multiple Choice

$$2x^{2\ }-11x+5$$

We rewrote the quadratic equation as

$$\left(2x^{2\ }-10x\right)\ +\left(-1x+5\right)$$

STEP 7: Factor each of the parentheses

• $$2x\left(x\ -5\right)\ -1\left(x-5\right)$$

• $$5x\left(x\ -2\right)\ +1\left(x+5\right)$$

• $$2x\left(x\ -5\right)\ +2x\left(x-5\right)$$

Multiple Choice

$$2x^{2\ }-11x+5$$

We rewrote the quadratic equation as

$$2x\left(x\ -5\right)\ -1\left(x-5\right)$$

STEP 8: Factor the (x-5) out

• $$\left(x-5\right)\left(2x\ -1\right)$$

• $$\left(2x\ -1\right)$$

• $$\left(x-5\right)\left(2x\right)$$

Open Ended

$$2x^{2\ }-11x+5$$

We rewrote the quadratic equation as

$$\left(x-5\right)\left(2x\ -1\right)$$

STEP 9: Try to FOIL it back out to make sure you got it right

Multiple Select

The final factored form of the quadratic equation was:

$$\left(x-5\right)\left(2x\ -1\right)$$ .

Can you find the values of x ?

[HINT: set each set of parentheses equal to zero and then solve]

• x = 5

• x = $$\frac{1}{2}$$

• x = -1

• x = -5

• x = $$-\frac{1}{2}$$

Multiple Choice

Standard form of a polynomial is $$ax^2+bx+c=0$$

Let's try to do another example step by step:

$$10x^{2\ }-3x-1$$

STEP1: Identify a, b, and c.

• a = 10

  b = -3

  c = -1

• a = -10

  b = -3

  c = -1

• a = 10

  b = -3

  c = 1

• a = 10

  b = 1

  c = -1

• a = -10

  b = -3

  c = 3

Multiple Choice

$$10x^{2\ }-3x-1$$

a = 10

b = -3

c = -1

STEP 2: Multiply a and c to get our k

• 10(-1) = -10

  so k = -10

• 10(-1) = -3

  so k = -3

• 10(-1) = 13

  so k = 13

Multiple Select

$$10x^{2\ }-3x-1$$

k = 10(-1) = -10

STEP 3: Find all the factor pairs of k

• -1(10)

  and

  1(-10)

• -2(5)

  and

  2(-5)

• -3(3)

• 2(-8)

  ana

  -2(8)

Multiple Choice

$$10x^{2\ }-3x-1$$

the factor pairs of k are

-1(10) and 1(-10) and 2(-5) and -2(5)

STEP 4: which one of these factor pairs add up to b = -3

• -1(10)

  because

  -1 + 10 = -3

• 1(-10)

  because

  -10 + 1 = -3

• -2(5)

  because

  -2 + 5 = -3

• 2(-5)

  because

  -5 + 2 = -3

Multiple Choice

$$10x^{2\ }-3x-1$$

the factor pair we picked was -5(2) because

-5 + 2 = -3, which is our b constant.

STEP 5: rewrite our quadratic expression using -5 and 2 instead of -3

• $$10x^{2\ }-5x\ +2x-1$$

• $$10x^{2\ }+5x\ +2x-1$$

• $$10x^{2\ }-5x\ -2x-1$$

• $$10x^{2\ }+5x\ -2x-1$$

Multiple Choice

$$10x^{2\ }-3x-1$$

We rewrote the quadratic equation as

$$10x^{2\ }-5x\ +2x-1$$

STEP 6: separate our rewritten quadratic equation using two sets of parentheses.

e.g. (something) + (something)

• $$\left(10x^{2\ }-5x\right)\ +\left(2x-1\right)$$

• $$\left(10x^2+\left(-5x+2x\right)-1\right)$$

Multiple Choice

$$10x^{2\ }-3x-1$$

We rewrote the quadratic equation as

$$\left(10x^{2\ }-5x\right)\ +\left(2x-1\right)$$

STEP 7: Factor each of the parentheses

• $$5x\left(2x\ -1\right)\ +1\left(2x-1\right)$$

• $$2x\left(2x\ -1\right)\ +1\left(2x-1\right)$$

• $$5x\left(x\ -5\right)\ +1\left(2x-1\right)$$

Multiple Choice

$$10x^{2\ }-3x-1$$

We rewrote the quadratic equation as

$$5x\left(2x\ -1\right)\ +1\left(2x-1\right)$$

STEP 8: Factor out the (2x-1)

• $$\left(5x+1\right)\left(2x\ -1\right)$$

• $$\left(2x\ -1\right)$$

• $$\left(5x+1\right)\left(2x\right)$$

Open Ended

$$10x^{2\ }-3x-1$$

We rewrote the quadratic equation as

$$\left(5x+1\right)\left(2x\ -1\right)$$

STEP 9: Try to FOIL it back out to make sure you got it right

Multiple Select

The final factored form of the quadratic equation was

$$\left(5x+1\right)\left(2x\ -1\right)$$ .

Can you find the values of x?

[HINT: set each set of parentheses equal to zero and then solve]

• x = $$-\frac{1}{5}$$

• x = $$\frac{1}{2}$$

• x = $$-\frac{1}{2}$$

• x = $$\frac{1}{5}$$

• x = -5

Practice it

Multiple Choice

Factor the quadratic equation:

$$9x^2-12x-5$$

• ( 3x - 5 )( 3x + 1 )

• ( 3x - 5 )( 3x + 5 )

• ( 3x - 5 )( 3x - 1 )

• ( 3x + 5 )( 3x + 1 )

• ( 3x + 5 )( 3x - 1 )

Multiple Choice

Factor the quadratic equation:

$$4x^2+37x+9$$

• ( 2x + 1)( 2x + 3 )

• ( 4x + 3 )( x + 3 )

• ( 2x + 1 )( 2x + 9 )

• ( 4x + 1 )( x + 3 )

• ( 4x + 1 )( x + 9 )

Multiple Choice

Factor the quadratic equation:

$$3x^2+11x\ -4$$

• ( 3x - 1 )( x + 4 )

• ( 3x - 1 )( x - 4 )

• ( 3x + 1 )( x + 4 )

• ( 3x - 4 )( x + 1 )

• ( 3x + 1 )( x - 4 )

Multiple Choice

Factor the quadratic equation:

$$10x^2+33x-7$$

• ( 2x + 1 )( 5x - 7 )

• ( 2x + 1 )( 5x + 7 )

• ( 2x - 1 )( 5x + 7 )

• ( 2x - 1 )( 5x - 7 )

• ( 10x - 1 )( x + 7 )

Multiple Choice

Factor the quadratic equation:

$$10x^2-3x-1$$

• ( 10x - 1 )( x + 1 )

• ( 2x - 1 )( 5x - 1 )

• ( 2x - 1 )( 5x + 1 )

• ( 2x + 1 )( 5x + 1 )

• ( 2x + 1 )( 5x - 1 )

Multiple Choice

Factor the quadratic equation:

$$4x^2+19x-30$$

• ( 2x - 5 )( 2x + 6 )

• ( 4x + 5 )( x - 6 )

• ( 4x + 5 )( x + 6 )

• ( 4x - 5 )( x - 6 )

• ( 4x - 5 )( x + 6 )