How to Easily Solve Math Problems Using Difference of Squares
Set up a difference of squares formula., Subtract one factor from the other., Divide the difference between the two factors by 2., Find the number equidistant from the two factors., Plug the equidistant number into the difference of squares...
Step-by-Step Guide
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Step 1: Set up a difference of squares formula.
The difference of squares formula states that (x−y)(x+y)=x2−y2{\displaystyle (x-y)(x+y)=x^{2}-y^{2}}.
In the formula, x{\displaystyle x} = the number equidistant from the two factors, and y{\displaystyle y} = half the difference between the two factors., Determine whether their difference is even or odd.
If their difference is even, that means they are equidistant from another number, and you can use the difference of squares method.A factor is a number being multiplied by another number.
If the difference between factors is odd, then you can still use the difference of squares method to solve, but you will have to manipulate the problem, as described in Method
2.
For example, if you were multiplying 28×32{\displaystyle 28\times 32}, you could use the difference of squares method, because 28−32=−4{\displaystyle 28-32=-4}, and
-4 is an even number.
It doesn’t matter if the difference is positive or negative.
A difference of
-4 is the same as a difference of
4.
Both +/-4 are even. , Plug in this number for y{\displaystyle y} in the difference of squares formula.
For example, you already found that the difference between 28 and 32 is
4. 4÷2=2{\displaystyle 4\div 2=2}.
So half the difference between the two factors is
2.
Plugging in this number for y{\displaystyle y}, your formula becomes (x−2)(x+2)=x2−22{\displaystyle (x-2)(x+2)=x^{2}-2^{2}}. , To do this, take the larger factor and subtract it by half the difference between the two factors (y{\displaystyle y} in your difference of squares formula).
For example, if the larger factor is 32, and half the difference between factors (y{\displaystyle y}) is 2, then you would find the equidistant number by subtracting 2 from
32. 32−2=30{\displaystyle 32-2=30}, so the equidistant number is
30.
You could also find the equidistant number by taking the smaller factor and adding half the difference. , To do this, replace all the x{\displaystyle x}s in the formula with the equidistant number.
For example, if your equidistant number is 30, your difference of squares formula will look like this: (30−2)(30+2)=302−22{\displaystyle (30-2)(30+2)=30^{2}-2^{2}}.
Note that (30−2)(30+2){\displaystyle (30-2)(30+2)} is the same as (28)(32){\displaystyle (28)(32)}, which was your original problem. , Plug these squares into the equation.
Remember, to square something means to multiply it by itself.
For example:302−22{\displaystyle 30^{2}-2^{2}}=900−4{\displaystyle 900-4} , The result will equal the answer to your original multiplication problem.
For example, 900−4=896{\displaystyle 900-4=896}, so 28×32=896{\displaystyle 28\times 32=896}. , You can verify your answer by using a calculator, or by solving using the standard multiplication algorithm. -
Step 2: Subtract one factor from the other.
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Step 3: Divide the difference between the two factors by 2.
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Step 4: Find the number equidistant from the two factors.
-
Step 5: Plug the equidistant number into the difference of squares formula.
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Step 6: Square the two numbers.
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Step 7: Complete the subtraction.
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Step 8: Check your work.
Detailed Guide
The difference of squares formula states that (x−y)(x+y)=x2−y2{\displaystyle (x-y)(x+y)=x^{2}-y^{2}}.
In the formula, x{\displaystyle x} = the number equidistant from the two factors, and y{\displaystyle y} = half the difference between the two factors., Determine whether their difference is even or odd.
If their difference is even, that means they are equidistant from another number, and you can use the difference of squares method.A factor is a number being multiplied by another number.
If the difference between factors is odd, then you can still use the difference of squares method to solve, but you will have to manipulate the problem, as described in Method
2.
For example, if you were multiplying 28×32{\displaystyle 28\times 32}, you could use the difference of squares method, because 28−32=−4{\displaystyle 28-32=-4}, and
-4 is an even number.
It doesn’t matter if the difference is positive or negative.
A difference of
-4 is the same as a difference of
4.
Both +/-4 are even. , Plug in this number for y{\displaystyle y} in the difference of squares formula.
For example, you already found that the difference between 28 and 32 is
4. 4÷2=2{\displaystyle 4\div 2=2}.
So half the difference between the two factors is
2.
Plugging in this number for y{\displaystyle y}, your formula becomes (x−2)(x+2)=x2−22{\displaystyle (x-2)(x+2)=x^{2}-2^{2}}. , To do this, take the larger factor and subtract it by half the difference between the two factors (y{\displaystyle y} in your difference of squares formula).
For example, if the larger factor is 32, and half the difference between factors (y{\displaystyle y}) is 2, then you would find the equidistant number by subtracting 2 from
32. 32−2=30{\displaystyle 32-2=30}, so the equidistant number is
30.
You could also find the equidistant number by taking the smaller factor and adding half the difference. , To do this, replace all the x{\displaystyle x}s in the formula with the equidistant number.
For example, if your equidistant number is 30, your difference of squares formula will look like this: (30−2)(30+2)=302−22{\displaystyle (30-2)(30+2)=30^{2}-2^{2}}.
Note that (30−2)(30+2){\displaystyle (30-2)(30+2)} is the same as (28)(32){\displaystyle (28)(32)}, which was your original problem. , Plug these squares into the equation.
Remember, to square something means to multiply it by itself.
For example:302−22{\displaystyle 30^{2}-2^{2}}=900−4{\displaystyle 900-4} , The result will equal the answer to your original multiplication problem.
For example, 900−4=896{\displaystyle 900-4=896}, so 28×32=896{\displaystyle 28\times 32=896}. , You can verify your answer by using a calculator, or by solving using the standard multiplication algorithm.
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Shirley Lewis
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