(2m2+10m+1) from (−8m2−8m+12)

4mfro• (2m2+10m+1)•(2m2+2m-3)

STEP

1

:

Equation at the end of step 1

0-2•(m2))+10m)+1)•f)•r)•o)•m)•(((0-23m2)-8m)+12))

STEP

2

:

Equation at the end of step

2

:

0-2m2+10m)+1)•f)•r)•o)•m)•(-8m2-8m+12))

STEP

3

:

Trying to factor by splitting the middle term

3.1 Factoring 2m2+10m+1

The first term is, 2m2 its coefficient is 2 .

The middle term is, +10m its coefficient is 10 .

The last term, "the constant", is +1

Step-1 : Multiply the coefficient of the first term by the constant 2 • 1 = 2

Step-2 : Find two factors of 2 whose sum equals the coefficient of the middle term, which is 10 .

-2 + -1 = -3

-1 + -2 = -3

1 + 2 = 3

2 + 1 = 3

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step

3

:

0-f•(2m2+10m+1)•r)•o)•m)•(-8m2-8m+12))

STEP

4

:

Equation at the end of step 4

0-(((fr•(2m2+10m+1)•o)•m)•(-8m2-8m+12))

STEP

5

:

Equation at the end of step 5

0-((fro•(2m2+10m+1)•m)•(-8m2-8m+12))

STEP

6

:

Equation at the end of step 6

0 - (mfro • (2m2 + 10m + 1) • (-8m2 - 8m + 12))

STEP

7

:

STEP

8

:

Pulling out like terms

8.1 Pull out like factors :

-8m2 - 8m + 12 = -4 • (2m2 + 2m - 3)

Trying to factor by splitting the middle term

8.2 Factoring 2m2 + 2m - 3

The first term is, 2m2 its coefficient is 2 .

The middle term is, +2m its coefficient is 2 .

The last term, "the constant", is -3

Step-1 : Multiply the coefficient of the first term by the constant 2 • -3 = -6

Step-2 : Find two factors of -6 whose sum equals the coefficient of the middle term, which is 2 .

-6 + 1 = -5

-3 + 2 = -1

-2 + 3 = 1

-1 + 6 = 5

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step

8

:

0 - -4mfro • (2m2 + 10m + 1) • (2m2 + 2m - 3)

STEP

9

:

Final result :

4mfro • (2m2 + 10m + 1) • (2m2 + 2m - 3)

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.