Factoring Trinomials with a = 1 04:26 minutes
Transcript Factoring Trinomials with a = 1
Captain Johnny Redbeard and his parrot Holly are hunting for treasure. Redbeard has heard many stories about the riches to be found on Treasure Island. Legend says that there's a chest no man can open. It has a lock impervious to tools of all kinds and only the right combination of numbers will reveal the treasure hidden inside.
Factoring Trinomials
Eager to take on the challenge, Redbeard quickly sets course for Treasure Island. Taking a closer look at the map, Redbeard, sometimes referred to as the math expert of the seven seas, realizes that the treasure is not only marked by x, but also with a polynomial. He knows just what to do, he needs to factor the trinomial.
To open the treasure chest, Captain Johnny Redbeard has to find the correct set of binomials terms that, when multiplied, equal the the expression on the map, x²+6x27. So he needs to factor the expression.A standard trinomial is in the form ax² + bx + c. In this puzzle, a = 1, b = 6 and c = 27."
Reverse FOIL Method
Captain Johnny knows the reverse FOIL method to factor trinomials. Let's think of the two binonmials we're looking for as (x+m)(x+n). Remember, if we multiply these binomials using FOIL, we get x²+nx+mx+mn. If we compare this expression with the standard form, 'b' must be the sum of n+m whereas 'c' must be the product of 'm' and 'n'. Remember, you can use the Commutative Property for sums and products.
So, Captain Johnny Redbeard has to find exactly two numbers that follow these rules: The sum of the factors has to be 'b', in this case, 6. And also, the product of the same two numbers has to be 'c', in this case 27. There can't be that many options, but how exactly can we figure out the right numbers? Because there are fewer options for multiplication, let's start with that. Let's think of the possible factors of 27. The factors we have are 1 times (27), 1 times (27), 3 times (9), and 3 times (9). These are the only possible values for 'm' and 'n' here.
Let's look at the sum of these factors. Remember, we're looking for 6 as the result. Both 1 + 27 and 1 + 27 are not equal to 6. 3 + 9 = 6 also not equal to 6. So now we check our last combination 3 + 9 = 6. We've found our correct 'm' and 'n' values! We know that 3 and 9 are our 'm' and 'n' values, where m = 3 and n = 9. We can plug these into our parentheses as the factors of x² + 6x  27. Our factors are (x  3) and (x + 9).
Well, shiver me timbers! Captain Johnny Redbeard managed to solve the riddle, which no man was able to do before. As he enters the factors into lock on the treasure chest the treasure chest reveals. BLIMEY!!! another map with more difficult looking expressions but at least he's good at math! Weigh anchor and hoist the mizzen! ...er...something like that...anyway, he's off on another adventure!

Introduction to Polynomials – Naming Polynomials by Number of Terms

Adding Polynomials

Multiplying Polynomials

Multiplying Special Case Polynomials

Factoring out the GCF

Factoring Trinomials with a = 1

Factoring Trinomials with a ≠ 1

Factoring Special Case Polynomials

Factoring by Grouping

Subtracting Polynomials