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}

Algebra Honors ( Period 6 & 7)

Squares of Binomials 5-6

(a +b)² =(a+b)(a+b)

You could use foil, fireworks, the box method… to find the results of multiplying the two binomials

a² + 2ab + b²

What happens when you square the binomial difference a-b?

(a-b)² = (a-b)(a-b)

a² - 2ab + b²

Notice you have the square of the first term,  twice the product of the two terms and finally the square of the last term.

The textbook states that it is helpful to memorize these patterns for writing squares of binomials as trinomials.

(a +b)²     = a² + 2ab + b²

(a - b)²     = a² -2ab + b²

My comment-  MEMORIZE … you need to be able to see these patterns

(x + 3)² = x² + 6x + 9

( 7u -3)² = 49u² -42u+ 9 

(4s – 5t)² = 16s² – 40st + 25t²

(3p² – 2q²)²

=9p⁴-12p²q² + 4q⁴

When we need to factor… we need to realize these patterns in reverse…that is,

a² + 2ab + b^{2  =}    (a +b)²     

a² -2ab + b²   =  (a - b)²     

a² + 2ab + b²     and  a² -2ab + b²       are called perfect square trinomials

because each expression has a three terms and is the square of a binomial.

To test whether a trinomial is a perfect square… ask these three questions:

1)  Is the first term a square?

2) Is the last term a square?

3) Is the middle term twice the product of √( 1^st term)  and √(last term).

For example, is

4x² – 20x + 25 a perfect square trinomial?

 1)  Is the first term a square? YES 4x² = (2x)²

2) Is the last term a square? YES  25= (5)²

3) Is the middle term twice the product of √( 1^st term)  and √(last term).

YES  2[√( 4x²) ∙ √25)]. =  2(2x∙5) = 20x

Yu may need to rearrange the terms of a trinomial BEFORE you test whether it is a perfect square.

For example, x² + 100 -20x must be rewritten as

x² -20x + 100  so that we can answer YES to all three questions.

Take a look at

63n³ – 84n² + 28n

It doesn’t look like a perfect square BUT if you factor out the GCF or the Greatest Monomial factor…

we end up with à 7n(9n² -12n +4)

Now… that becomes  7n(3n-2)²

What about,

8u3 -24u2v + 18uv2

2u(4u² -12vu +9v²)

2u(2u-3v)²

What do we need to do to solve the following:
(x + 2)² – (x -3) ² = 35

First multiply each of the binomial squares

x² + 4x + 4 – (x² -6x+ 9) = 35

Make sure to properly employ the inverse property of a sum

x² + 4x + 4 – x² + 6x -  9 = 35

Combine like terms

10x – 5 = 35

10x = 40

x = 4

or using set notation {4}

Math 7 (Period 4)

I just found these cute You Tube Videos about Integers

Integers


Adding Negative Integers


Adding Negative Integers Quiz