Algebra Honors ( Periods 6 & 7)

Factoring Pattern for x² + bx+ c  where c is  positive 5-7 

In this lesson we will be factoring trinomials that can be factored as a product of ( x +r)(x + s)
where r and s are both positive OR both negative integers.

x² + ( r + s)x + rs  

(x +3)(x+5) = x² + 8x + 15

(x – 6)(x -4) = x² -10x + 24

where -10 is the sum of -6 and -4

and

24 is the product of -6 and -4

Our book suggests that you list all the pairs of integral factors whose products equal the constant term Then, find the pair of integral factors whose SUM equals the coefficient of the linear term (remember your new vocab)

For factoring x² + bx + c, where c is positive  you only need to consider factors WITH THE SAME SIGNS as the linear term!!

In class, I showed the diamond method of calculating products and sums… and even recommended an app to practice!!
Here is the link for the iphone,  ipad app…

y² + 14y + 40

Since the linear term ( +14y) is positive you know to set up two sets of HUGS with + in the middle

(  +  )(  +  )

then you can add the single y’s since y² = y·y

(y + )(y + )

Now either list all the integral factors or use the diamond method and you discover that 10 and 4 are the two factors that multiply to 40 AND also sum to 14

(y + 10 )(y + 4)

To check if you are accurate, use FOIL or Fireworks… or the BOX method and see if you get back to the original problem!

y2 – 11y + 18

This time notice that the linear term is -11y so you will be looking for a pair of numbers whose product will be positive  ( so both need to be negative)

Set  up your  HUGS  similarly—EXCEPT both signs need to be NEGATIVE

(y -   )( y -  )

Since -11 is negative, think of the negative factors of 18 using the book’s method or the diamond method

hmmm… -9 and -2 work

(y - 9  )( y -  2)

Again To check if you are accurate, use FOIL or Fireworks… or the BOX method and see if you get back to the original problem!

A polynomial that cannot be expressed as a product of polynomials of lower degree is said to be irreducible. An irreducible polynomial with integral coefficients whose greatest monomial factor is 1 is a PRIME POLYNOMIAL.

Factor x² -10x + 14

Setting up your HUGS   you start to think what two factors multiply to 14 and SUM to 10… hmmm. NOTHING…

Therefore x² -10x + 14  cannot be factored and it is a prime polynomial

Find all the integral values of k for which the trinomial can be factored

x² + kx + 28

28 can be factored as a product—using  a T chart

list all the factors

(1)(28) (2)(14) (4)(7)

Taking the corresponding sums you get 29, 16 and 11

BUT… remember you can also have the negatives here

so the values of k can be ± 29,   ± 16,   ± 11

{-29, -16, -11, 11, 16, 29}