SAT Math › Factoring Equations
Factor the following equation.
x2 – 16
(x + 4)(x + 4)
(x – 4)(x – 4)
(x + 4)(x – 4)
(x)(x – 4)
(x2)(4 – 2)
The correct answer is (x + 4)(x – 4)
We neen to factor x2 – 16 to solve. We know that each parenthesis will contain an x to make the x2. We know that the root of 16 is 4 and since it is negative and no value of x is present we can tell that one 4 must be positive and the other negative. If we work it from the multiple choice answers we will see that when multiplying it out we get x2 + 4x – 4x – 16. 4x – 4x cancels out and we are left with our answer.
Factor the following equation.
x2 – 16
(x + 4)(x + 4)
(x – 4)(x – 4)
(x + 4)(x – 4)
(x)(x – 4)
(x2)(4 – 2)
The correct answer is (x + 4)(x – 4)
We neen to factor x2 – 16 to solve. We know that each parenthesis will contain an x to make the x2. We know that the root of 16 is 4 and since it is negative and no value of x is present we can tell that one 4 must be positive and the other negative. If we work it from the multiple choice answers we will see that when multiplying it out we get x2 + 4x – 4x – 16. 4x – 4x cancels out and we are left with our answer.
if x – y = 4 and x2 – y = 34, what is x?
12
9
15
6
10
This can be solved by substitution and factoring.
x2 – y = 34 can be written as y = x2 – 34 and substituted into the other equation: x – y = 4 which leads to x – x2 + 34 = 4 which can be written as x2 – x – 30 = 0.
x2 – x – 30 = 0 can be factored to (x – 6)(x + 5) = 0 so x = 6 and –5 and because only 6 is a possible answer, it is the correct choice.
if x – y = 4 and x2 – y = 34, what is x?
12
9
15
6
10
This can be solved by substitution and factoring.
x2 – y = 34 can be written as y = x2 – 34 and substituted into the other equation: x – y = 4 which leads to x – x2 + 34 = 4 which can be written as x2 – x – 30 = 0.
x2 – x – 30 = 0 can be factored to (x – 6)(x + 5) = 0 so x = 6 and –5 and because only 6 is a possible answer, it is the correct choice.
Solve for a.
No solution
The expression can be factored.
We must find two numbers that added together equal -7, and multiplied together equal 60. Those two numbers are -12 and 5.
Now we can set both terms equal to zero and solve for a.
,
If x_2 + 2_ax + 81 = 0. When a = 9, what is the value of x?
0
3
9
–9
–18
When a = 9, then x_2 + 2_ax + 81 = 0 becomes
x_2 + 18_x + 81 = 0.
This equation can be factored as (x + 9)2 = 0.
Therefore when a = 9, x = –9.
If x_2 + 2_ax + 81 = 0. When a = 9, what is the value of x?
0
3
9
–9
–18
When a = 9, then x_2 + 2_ax + 81 = 0 becomes
x_2 + 18_x + 81 = 0.
This equation can be factored as (x + 9)2 = 0.
Therefore when a = 9, x = –9.
Solve for a.
No solution
The expression can be factored.
We must find two numbers that added together equal -7, and multiplied together equal 60. Those two numbers are -12 and 5.
Now we can set both terms equal to zero and solve for a.
,
Factor .
Cannot be factored
First pull out any common terms: 4_x_3 – 16_x_ = 4_x_(_x_2 – 4)
_x_2 – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is _a_2 – _b_2 = (a – b)(a + b). Here a = x and b = 2. So _x_2 – 4 = (x – 2)(x + 2).
Putting everything together, 4_x_3 – 16_x_ = 4_x_(x + 2)(x – 2).
Factor .
Cannot be factored
First pull out any common terms: 4_x_3 – 16_x_ = 4_x_(_x_2 – 4)
_x_2 – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is _a_2 – _b_2 = (a – b)(a + b). Here a = x and b = 2. So _x_2 – 4 = (x – 2)(x + 2).
Putting everything together, 4_x_3 – 16_x_ = 4_x_(x + 2)(x – 2).