Given the polynomial expression \( x^2 - 4x - 12 \), Alvin thinks that an equivalent expression is \( (x + 3)(x - 4) \). Simon thinks that an equivalent expression is \( (x -12)(x + 1) \). Theodore thinks that an equivalent expression is \( (x + 2)(x - 6) \) . Dave thinks that Alvin, Simon, and Theodore are all correct because the product of \( (3)(-4) = -12, (-12)(1) = -12 \), and \( (+2)(-6)=-12 \). Whom do you agree with and why?
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We are going to learn how to factor polynomials. Remember from target B which dealt with factoring a GCF, when we "factor" we are breaking up an expression into new expressions that when multiplied together are equivalent to the original expression. We will start with factoring two special polynomials that we learned about in Polynomial Target D, difference of two squares and perfect square trinomials, then we will move on to a basic trinomial that has a leading coefficient of \(1\) of the form: \( x^2 + bx + c \), next we will look at a trinomial that has a leading coefficient other than 1 of the form: \( ax^2 + bx + c \) finally we factor completely. Let's begin.
Factoring: Difference of Two Squares
Factoring: Difference of Two Squares
Factoring: Perfect Square Trinomial
Factoring a trinomial of the form \(x^2 + bx +c \)
Let's see what it looks like to factor a trinomial of the form \( x^2 + bx + c \) (notice that the leading coefficient is \(1\)).
Let's see what it looks like to factor a trinomial of the form \( x^2 + bx + c \) (notice that the leading coefficient is \(1\)).
The video taught us a few key points:
1. We see that in order to factor a trinomial like \( x^2 + bx + c \) it will be factored into \(2\) binomials that are being multiplied together.
2. We begin to see where the terms in the binomials come from (the reverse of when we are multiplying).
1. We see that in order to factor a trinomial like \( x^2 + bx + c \) it will be factored into \(2\) binomials that are being multiplied together.
2. We begin to see where the terms in the binomials come from (the reverse of when we are multiplying).
3. This idea of product and sum is introduced. This form of "guess and check" using the idea of product sum works when the leading coefficient is one. When we get to other trinomials with a leading coefficient other than one, the guess and check becomes a bit more challenging.
Let's now factor a trinomial \( x^2 + bx + c \) with leading coefficient of \(1\).
Let's now factor a trinomial \( x^2 + bx + c \) with leading coefficient of \(1\).
Factoring a trinomial \(ax^2 + bx +c \)
Now we are going to factor a trinomial when the leading coefficient other than 1. We will still use the idea of "guess and check" but we will see how the \(a\) value will now play a part. Remember this is a trinomial so we will factoring or writing it in an equivalent form of the multiplication of two binomials.
Now we are going to factor a trinomial when the leading coefficient other than 1. We will still use the idea of "guess and check" but we will see how the \(a\) value will now play a part. Remember this is a trinomial so we will factoring or writing it in an equivalent form of the multiplication of two binomials.
As you can see from the previous examples, the "guess and check" idea really plays a part when we are factoring a trinomial of the form \( ax^2 + bx + c \).
Factoring a trinomial \(ax^2 + bx +c \) (By Grouping)
Above we could see that guessing and checking possible factors is a valid way to factor a trinomial, regardless of the value of the leading coefficient. Another process that can be used is called Factoring By Grouping. To do this, we will still apply the idea of product and sum, but the leading coefficient will play a larger role in this thought process.
Just like the previous two videos, our two factors will result in two binomials. Further, we will see our multiplication of those binomials in reverse order after we are done with all our work.
Just like the previous two videos, our two factors will result in two binomials. Further, we will see our multiplication of those binomials in reverse order after we are done with all our work.
Factoring Completely
When we factor completely, it means we write an equivalent expressions of the polynomial with factors that are multiplied together and there may be more than "2" factors multiplied together. If we have more than two factors that means we may need to use more than one type of factoring. When we begin to factor, we will always look for a GCF first, then from there we will look to identify if it is a difference of two squares or a perfect square trinomial. If it isn't either of these two, we will then go to guess and check to factor. The following is a basic check list to help remind you in which order to check to factor.
When we factor completely, it means we write an equivalent expressions of the polynomial with factors that are multiplied together and there may be more than "2" factors multiplied together. If we have more than two factors that means we may need to use more than one type of factoring. When we begin to factor, we will always look for a GCF first, then from there we will look to identify if it is a difference of two squares or a perfect square trinomial. If it isn't either of these two, we will then go to guess and check to factor. The following is a basic check list to help remind you in which order to check to factor.
So when we factor completely we need to make sure to check for a GCF first and then continue through our checklist. Just because we are factoring completely doesn't necessarily mean we will have to factor more than once. It is just to make sure we are writing the equivalent expression in the most basic form.
Quick Check
Write an equivalent expression by factoring the difference of two squares.
1) \( x^2 - 81 \)
Write an equivalent expression by factoring the perfect square trinomial.
2) \( x^2 - 20x +100 \)
Write an equivalent expression by factoring the trinomial.
3) \( x^2 -3x - 10 \)
Write an equivalent expression by factoring completely.
4) \( 3x^2 - 18x - 48 \)
5) \( 2x^2 - 98 \)
Quick Check Solutions
Write an equivalent expression by factoring the difference of two squares.
1) \( x^2 - 81 \)
Write an equivalent expression by factoring the perfect square trinomial.
2) \( x^2 - 20x +100 \)
Write an equivalent expression by factoring the trinomial.
3) \( x^2 -3x - 10 \)
Write an equivalent expression by factoring completely.
4) \( 3x^2 - 18x - 48 \)
5) \( 2x^2 - 98 \)
Quick Check Solutions