Chapter Closure
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Exercise 134 Page 252
a
Is there a greatest common factor (GCF) between all of the terms in the given expression? If so, you should factor that out first.
b
Is there a GCF between all of the terms in the given expression? If so, you should factor that out first.
c
Is there a GCF between all of the terms in the given expression? If so, you should factor that out first.
d
Is there a GCF between all of the terms in the given expression? If so, you should factor that out first.
a
(x+2)(3x+5)
b
x(x-5)(6x-1)
c
3a(b-1)(2b+7)
d
(y-3)(y+8)
a
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.
3x^2+11x+10
We have that a= 3, b=11, and c=10. There are now three steps we need to follow in order to rewrite the above expression.
- Find a c. Since we have that a= 3 and c=10, the value of a c is 3* 10=30.
- Find factors of a c. Since ac=30, which is positive, we need factors of a c to have the same sign in order for the product to be positive. Since b=11, which is also positive, those factors will need to be positive so that their sum is positive.
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result 1 &30 &1 + 30 &31 2 &15 &2 + 15 &17 3 &10 &3 + 10 &13 5 & 5 & 5 + 6 &11
- Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 3x^2+11x+10 ⇕ 3x^2+ 6x+ 5x+10
Finally, we will factor the last expression obtained.
b
We want to completely factor the given expression. To do so, we will first identify and factor out the greatest common factor.
Factor Out the GCF
The greatest common factor (GCF) of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. The GCF of the given expression is x.
6x^3-31x^2+5x
SplitIntoFactors
Split into factors
x(6x^2)- x(31x)+ x(5)
FactorOut
Factor out x
x(6x^2-31x+5)
Factor the Quadratic Trinomial
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. x( 6x^2-31x+5 ) ⇔ x( 6x^2+(-31)x+5 ) We have that a= 6, b=-31, and c=5. 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=5, the value of a c is 6* 5=30.
- Find factors of a c. Since a c=30, which is positive, we need factors of a c to have the same sign in order for the product to be positive. Since b=-31, which is negative, those factors will need to be negative so that their sum is negative.
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result 1 &30 &1 + 30 &31 -1 & -30 & -1 + ( -30) &-31
- Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. x(6x^2+(-31)x+5 ) ⇕ x(6x^2 - 30x - 1x+5 )
Finally, we will factor the last expression obtained.
x(6x^2-30x-1x+5 )
FactorOut
Factor out 6x
x(6x(x-5)-1x+5 )
FactorOut
Factor out -1
x(6x(x-5)-1(x-5) )
FactorOut
Factor out (x-5)
x(x-5)(6x-1)
c
We want to completely factor the given expression. To do so, we will first identify and factor out the greatest common factor.
Factor Out the GCF
The greatest common factor (GCF) of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. The GCF of the given expression is 3a.
6ab^2+15ab-21a
SplitIntoFactors
Split into factors
3a(2b^2)+ 3a(5b)- 3a(7)
FactorOut
Factor out 3a
3a(2b^2+5b-7)
Factor the Quadratic Trinomial
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. 3a( 2b^2+5b-7 ) ⇔ 3a( 2b^2+5b+(-7) ) We have that a= 2, b=5, and c=-7. There are now three steps we need to follow in order to rewrite the above expression.
- Find a c. Since we have that a= 2 and c=-7, the value of a c is 2* -7=-14.
- Find factors of a c. Since ac=-14, which is negative, we need factors of a c to have opposite signs in order for the product to be negative. Since b=5, which is positive, the absolute value of the positive factor will need to be greater than the absolute value of the negative factor, so that their sum is positive.
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result -1 &14 &-1 + 14 &13 -2 & 7 & -2 + 7 &5
- Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 3a(2b^2+5b-7 ) ⇕ 3a ( 2b^2 - 2b+ 7b-7 )
Finally, we will factor the last expression obtained.
3a(2b^2-2b+7b-7 )
FactorOut
Factor out 2b
3a(2b(b-1)+7b-7 )
FactorOut
Factor out 7
3a(2b(b-1)+7(b-1) )
FactorOut
Factor out (b-1)
3a(b-1)(2b+7)
d
To factor a trinomial with a leading coefficient of one, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
y^2+5y- 24
In this case, we have -24. This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have the opposite sign (one positive and one negative.)
| Factor Constants | Product of Constants |
|---|---|
| 1 & -24 or -1 & 24 | -24 |
| 2 & -12 or -2 & 12 | -24 |
| 3 & -8 or -3 & 8 | -24 |
| 4 & -6 or -4 & 6 | -24 |
Next, let's consider the coefficient of the linear term. y^2+5y- 24 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 5.
| Factors | Sum of Factors |
|---|---|
| 1 and -24 | -23 |
| -1 and 24 | 23 |
| 2 and -12 | -10 |
| -2 and 12 | 10 |
| 3 and -8 | -5 |
| -3 and 8 | 5 |
We found the factors whose product is -24 and whose sum is 5. y^2+5y- 24 ⇔ (y-3)(y+8)
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