a Find factors or terms a and b, such that ab=-10x^2 and a+b=3x.
B
b Find factors or terms a and b, such that ab=-6x^2 and a+b=- x.
C
c Find factors or terms a and b, such that ab=12x^2 and a+b=13x.
D
d Find factors or terms a and b, such that ab=14x^2 and a+b= 5x.
E
e Find factors or terms a and b, such that ab=-10x^2 and a+b=3x.
F
f Find factors or terms a and b, such that ab=-10x^2 and a+b=3x.
A
a (x-1)(2x+5)
B
b (x-3)(x+2)
C
c (3x+1)(x+4)
D
d Cannot be factored. See solution.
E
e (3x-2)(x+4)
F
f 2(x+4)(3x+1)
Practice makes perfect
a Using the process from the Math Notes about factoring quadratic expressions, we know that 2x^2 and - 5 goes into the lower left and upper right corners of the generic rectangle.
To fill in the remaining two rectangles we need two x-terms that have a sum of 3x and a product of -10x^2.
Notice that the product is negative, which means one factor must be negative and the other positive. With this in mind, let us factor - 10x^2 in as many ways as we can and sum the factors.
|c|c|c|r|c|
[-1em]
Product & ax(bx) & ax+bx & Sum & 3x? [0.2em]
[-1em]
-10x^2 & -10x(x) & -10x+x& -9x & * [0.1em]
-10x^2 & -5x(2x) & -5x+2x& -3x & * [0.1em]
-10x^2 & -2x(5x) & -2x+5x& 3x & âś“ [0.1em]
When one term is -2x and the other is 5x, we have a product of -10x^2 and a sum of 3x. Now we can complete the diamond and generic rectangle.
To factor the quadratic expression we add each side of the generic rectangle and multiply the sums.
2x^2+3x-5=(x+(- 1))(2x+5)
⇓
2x^2+3x-5=(x-1)(2x+5)
b Like in Part A, we will create a generic rectangle and a diamond using the quadratic expression.
Notice that the product is negative, which means that one factor must be negative and the other positive. With this in mind, let us factor -6x^2 in as many ways as we can and sum the factors.
|c|c|c|c|c|
[-1em]
Product & ax(bx) & ax+bx & Sum & - x? [0.2em]
[-1em]
-6x^2 & -6x(x) & -6x+x& -5x & * [0.1em]
-6x^2 & -3x(2x) & -3x+2x& - x & âś“ [0.1em]
-6x^2 & -2x(3x) & -2x+3x& x & * [0.1em]
-6x^2 & - x(6x) & - x+6x& 5x & * [0.1em]
When one term is -3x and the other is 2x, we have a product of -6x^2 and a sum of - x. Now we can complete the diamond and the generic rectangle.
To factor the quadratic expression we add the parts along each side of the generic rectangle and multiply the sums.
x^2-x-6=(x+(-3))(x+2)
⇓
x^2-x-6=(x-3)(x+2)
c Like in Parts A and B, we will create a generic rectangle and a diamond using the quadratic expression.
Both the product and the sum are positive. For this to be true, both factors and terms must be positive. With this in mind, let us factor 12x^2 in as many ways as we can and sum the factors.
|c|c|c|r|c|
[-1em]
Product & ax(bx) & ax+bx & Sum & 13x? [0.2em]
[-1em]
12x^2 & 12x(x) & 12x+x& 13x & âś“ [0.1em]
12x^2 & 6x(2x) & 6x+2x & 8x & * [0.1em]
12x^2 & 4x(3x) & 4x+3x& 7x & * [0.1em]
When one factor is 12x and the other is x, we have a product of 12x^2 and a sum of 13x. Now we can complete the diamond and generic rectangle.
To factor the quadratic expression we add the parts along each side of the generic rectangle and multiply the sums.
3x^2+13x+4=(3x+1)(x+4)
d Like in previous parts, we will create a generic rectangle and a diamond using the quadratic expression.
Both the product and the sum are positive. For this to be true, both factors and terms must be positive. With this in mind, let us factor 14x^2 in as many ways as we can and sum the factors.
|c|c|c|r|c|
[-1em]
Product & ax(bx) & ax+bx & Sum & 5x? [0.2em]
[-1em]
14x^2 & 14x(x) & 14x+x& 15x & * [0.1em]
14x^2 & 7x(2x) & 7x+2x& 9x & * [0.1em]
As we can see, there is no pair of factors whose sum would be equal to 5x. As a result, our expression cannot be factored further.
e Before we create a generic rectangle and a diamond using the quadratic expression, we should notice that the terms in the expression have a common factor of 7. We need to factor it out first before proceeding.
7x^2-7x-42 ⇔ 7(x^2-x-6)
Now that the terms no longer have a common factor, we can create a generic rectangle and a diamond from the trinomial between the parentheses.
Notice that the product is negative, which means that one factor must be negative and the other positive. With this in mind, let us factor -6x^2 as much as we can and sum the factors.
|c|c|c|r|c|
[-1em]
Product & ax(bx) & ax+bx & Sum & - x? [0.2em]
[-1em]
-6x^2 & -6x(x) & -6x+x& -5x & * [0.1em]
-6x^2 & -3x(2x) & -3x+2x& - x & âś“ [0.1em]
-6x^2 & -2x(3x) & -2x+3x& x & * [0.1em]
-6x^2 & - x(6x) & - x+6x& 5x & * [0.1em]
To factor the quadratic expression we add the parts along each side of the generic rectangle and multiply the sums.
7(x^2-x-6)=7(x+(-3))(x+2)
⇓
7(x^2-x-6)=7(x-3)(x+2)
f Before we create a generic rectangle and a diamond using the quadratic expression, we should notice that terms in the expression have a common factor of 2. We need to factor it out first before proceeding.
6x^2+26x+8 ⇔ 2(3x^2+13x+4)
Now that terms no longer have a common factor, we can create a generic rectangle and a diamond from the trinomial between the parentheses.
Both the product and the sum are positive. For this to be true, both factors and terms must be positive. With this in mind, let us factor 13x^2 in as many ways as we can and sum the factors.
|c|c|c|r|c|
[-1em]
Product & ax(bx) & ax+bx & Sum & 13 x? [0.2em]
[-1em]
12x^2 & 12x(x) & 12x+x& 13x & âś“ [0.1em]
12x^2 & 6x(2x) & 6x+2x& 8x & * [0.1em]
12x^2 & 4x(3x) & 4x+3x& x & * [0.1em]
To factor the quadratic expression we add the parts along each side of the generic rectangle and multiply the sums.
2(3x^2+13x+4)=2(x+4)(3x+1)
