a To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
k^2-12k+20
In this case, we have 20. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign (both positive or both negative.)
Factor Constants
Product of Constants
1 and 20
20
-1 and -20
20
2 and 10
20
-2 and -10
20
4 and 5
20
-4 and -5
20
Next, let's consider the coefficient of the linear term.
k^2-12k+20
For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, -12.
Factors
Sum of Factors
1 and 20
21
-1 and -20
-21
2 and 10
12
-2 and -10
-12
4 and 5
9
-4 and -5
-9
We found the factors whose product is 20 and whose sum is -12.
k^2-12k+20 ⇔ (k-2)(k-10)
b Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠1 and there are no common factors. To factor this expression, we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b.
6x^2+17x-14 ⇔ 6x^2+17x+(- 14)
We have that a= 6, b=17, and c=- 14. There are now three steps we need to follow in order to rewrite the above expression.
Find a c. Since we have that a= 6 and c=- 14, the value of a c is 6* (- 14)=- 84.
Find factors of a c. Since ac=- 84, which is negative, we need factors of a c to have opposite signs — one positive and one negative. Since their sum b=17, which is positive, the absolute value of the positive factor will need to be greater than the absolute value of the negative factor.
c To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
x^2-8x+16
In this case, we have 16. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign (both positive or both negative.)
Factor Constants
Product of Constants
1 and 16
16
-1 and -16
16
2 and 8
16
-2 and -8
16
4 and 4
16
-4 and -4
16
Next, let's consider the coefficient of the linear term.
x^2-8x+16
For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, -8.
Factors
Sum of Factors
1 and 16
17
-1 and -16
-17
2 and 8
10
-2 and -8
-10
4 and 4
8
-4 and -4
-8
We found the factors whose product is 16 and whose sum is -8.
x^2-8x+16 ⇔ (x-4)(x-4)
Notice that both factors are the same, so we can rewrite the factored form using squares.
(x-4)(x-4) ⇔ (x-4)^2
d Here we have a quadratic trinomial of the form am^2+bm+c, where |a| ≠1 and there are no common factors. To factor this expression, we will rewrite the middle term, bm, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b.
9m^2-1 ⇔ 9m^2+ m+(- 1)
We have that a= 9, b= , and c=- 1. There are now three steps we need to follow in order to rewrite the above expression.
Find a c. Since we have that a= 9 and c=- 1, the value of a c is 9* (- 1)=- 9.
Find factors of a c. Since ac=- 9, which is negative, we need factors of a c to have opposite signs — one positive and one negative. Since their sum b= , both factor will need to have the same absolute value. In other words, those factors need to be opposite numbers.
e We want to decide what makes the polynomials from Parts A-D quadratic.
&k^2-12k+20
&6x^2+17x-14
&x^2-8x+16
&9m^2-1
Notice that the highest exponent of all these polynomials — and therefore their degree — is 2. A polynomial with a degree of 2 is called quadratic, regardless of number of terms.
Thus, although some of our polynomials are trinomials and some are binomials, they are all quadratic.
