Combining the Shortened Multiplication Formulas

Combining the square of a sum formulas

Meet the square of binomials formulas:

Product of the sum of two terms and their difference

$(X + Y)\cdot(X - Y) = X^2 - Y^2$

Click here to read more about Multiplication of the sum of two elements by the difference between them!

The difference of squares formula

$(X - Y)^2=X^2 - 2XY + Y^2$

Click here to read more about the formula for the difference of squares

The Square of a Sum Formula

$(X + Y)^2=X^2+ 2XY + Y^2$

Click here to read more about the formula for the Sum of Squares

Formulas relating to two expressions cubed

$(a+b)^3=a^3+3a^2 b+3ab^2+b^3$

$(a-b)^3=a^3-3a^2 b+3ab^2-b^3$

Click here to read more about Formulas for Cubic Expressions

Example

Practice an exercise that combines all the shortened multiplication formulas together:
$x^2+(5+x)(5-x)+x^2-6x+9-(6^3+3\cdot 6^2\cdot x+3\cdot 6\cdot x^2+x^3)=(2+x)^2+(6+x)^3$

Let's start from the beginning of the exercise. Observe that the expression
$(5+x)(5-x)=25-x^2$
matches the product of the sum of th two terms and their difference
$(X + Y)\cdot(X - Y) = X^2 - Y^2$
Let's proceed to the expression $x^2-6x+9$
We notice that it matches the difference of squares formula
$(X - Y)^2=X^2 - 2XY + Y^2$
However let's avoid touching it at this stage.
Let's continue to $6^3+3\cdot 6^2\cdot x+3\cdot 6\cdot x^2+x^3$
and we can see that it perfectly matches the formula for two terms cubed
$(a+b)^3=a^3+3a^2 b+3ab^2+b^3$
which means
$6^3+3\cdot 6^2\cdot x+3\cdot 6\cdot x^2+x^3=(6+x)^3$

Let's continue to the second side and observe that the expression $(2+x)^2$ matches the formula for sum of squares
$(X + Y)^2=X^2+ 2XY + Y^2$
Therefore
$(2+x)^2=4+4x+x^2$

Now let's insert the data:
$x^2+25-x^2+x^2-6x+9-(6+x)^3=4+4x+x^2+(6+x)^3$
Let's reduce the terms and solve as follows:
$-6x+34=4x+4$
Move the terms to opposite sides:
$8x=30$
$x=3.75$

Detailed explanation

Start practice

Test yourself on combining short multiplication formulas!

Solve the following equation:

$$ (x+2)^2=(2x+3)^2 $$

Practice more now

Combining the Shortened Multiplication Formulas

Let's recall the square of binomials formulas

Quadratic Identities

Product of the sum and difference of two terms
$(X + Y)\cdot(X - Y) = X^2 - Y^2$
You can use this formula when there is a product between the sum of two specific terms and the difference of those same terms.
Instead of showing them as a product of sum and difference, you can write $X^2 - Y^2$
This rule also works in reverse.

The Square of a Difference Formula
$(X - Y)^2=X^2 - 2XY + Y^2$
When we encounter two numbers with a minus sign between them - meaning a difference, and they are wrapped in parentheses and raised as one expression to the second power, we can use this formula.

The Square of a Sum Formula
$(X + Y)^2=X^2+ 2XY + Y^2$
- When encountering two numbers with a plus between them - meaning a sum, and they are both enclosed in parentheses and raised as one expression to the second power, we can use this formula.

Formulas relating to two expressions to the power of 3
$(a+b)^3=a^3+3a^2 b+3ab^2+b^3$
A way to express the sum of two terms, when they are enclosed in parentheses and raised as one expression to the power of three.
$(a-b)^3=a^3-3a^2 b+3ab^2-b^3$
A way to express the difference of two terms, when they are enclosed in parentheses and raised as one expression to the power of three.

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Test your knowledge

Question 1

Solve the following equation:

$$ (x-4)^2+3x^2=-16x+12 $$

Question 2

Find X

$$ 7x+1+(2x+3)^2=(4x+2)^2 $$

Question 3

Solve the following equation:

$$ (x+3)^2=2x+5 $$

Exercise

And now? Let's practice exercises that combine several perfect square formulas together!
Ready?
Here's an exercise:
Solve the following exercise using the perfect square formulas -
$x^2+4+(5x+1)^2=(4x-1)^2+(x+2)\cdot (x-2)$

Solution:
Although the exercise looks complex, if we solve it step by step using the square formulas, you'll see it's simple and easy to solve.
Let's start reading the exercise from left to right and understand which expression can match a square formula.
$(5x+1)^2$
This expression matches the square of sum formula -
The square of sum formula
$(X + Y)^2=X^2+ 2XY + Y^2$
Therefore
$(5x+1)^2=5^2+2\cdot 5x\cdot 1+1^2$
$(5x+1)^2=25x^2+10x+1$
Let's write this in the original exercise and we obtain:
$x^2+4+25x^2+10x+1=(4x-1)^2+(x+2)\cdot (x-2)$
Let's continue reading the exercise... aha
The expression
$(4x-1)^2$
matches the square of difference formula
The square of difference formula
$(X - Y)^2=X^2 - 2XY + Y^2$
Therefore
$(4x-1)^2=16x^2-8x+1$
Let's write this in the original exercise after we've dealt with it and obtain:
$x^2+4+25x^2+10x+1=16x^2-8x+1+(x+2)\cdot (x-2)$
Let's continue reading the exercise... aha
The expression
$(x+2)\cdot (x-2)$
matches the formula for the product of sum and difference
$(X + Y)\cdot(X - Y) = X^2 - Y^2$
Therefore we can replace
$(x+2)\cdot (x-2)$
with $x^2-2^2$
Let's write this in the original exercise after we've dealt with it:
$x^2+4+25x^2+10x+1=16x^2-8x+1+ x^2-2^2$

Note - we can write $2^2=4$ and obtain the following:
$x^2+4+25x^2+10x+1=16x^2-8x+1+ x^2-4$
Excellent! Notice that we can cancel $x^2-4$ from both sides and obtain:
$25x^2+10x+1=16x^2-8x+1$
Move terms to one side and obtain:
$9x^2+18x=0$
Factor out:
$9x(x^2+2)=0$
For the equation to equal zero, we can substitute $0$ once
because $9x=0$
$x=0$
And a second solution:
$(x^2+2)=0$
This expression will never equal zero because $x^2$ will never be negative, so there is only one solution which is $x=0$

Do you know what the answer is?

Question 1

Solve the equation

$$ 2x^2-2x=(x+1)^2 $$

Question 2

Solve the following equation:

$$ (x+2)^2=(2x+3)^2 $$

Question 3

Solve the following equation:

$$ (x-4)^2+3x^2=-16x+12 $$

Examples with solutions for Combining Short Multiplication Formulas

Exercise #1

Solve the following equation:

$(x+2)^2=(2x+3)^2$

Video Solution

Step-by-Step Solution

We will solve the equation $(x+2)^2 = (2x+3)^2$ by expanding and simplifying both sides:

Step 1: Expand both sides of the equation:
Left side: $(x+2)^2 = x^2 + 4x + 4$
Right side: $(2x+3)^2 = 4x^2 + 12x + 9$

Step 2: Set the expanded forms equal to each other:
$x^2 + 4x + 4 = 4x^2 + 12x + 9$

Step 3: Rearrange to form a standard quadratic equation:
Subtract $x^2 + 4x + 4$ from both sides:
$0 = 3x^2 + 8x + 5$

Step 4: Rearrange to get:
$3x^2 + 8x + 5 = 0$

Step 5: Solve using the quadratic formula:
Using $a = 3$ , $b = 8$ , $c = 5$ :
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3}$
$x = \frac{-8 \pm \sqrt{64 - 60}}{6}$
$x = \frac{-8 \pm \sqrt{4}}{6}$
$x = \frac{-8 \pm 2}{6}$

Step 6: Calculate the solutions:
$x_1 = \frac{-8 + 2}{6} = \frac{-6}{6} = -1$
$x_2 = \frac{-8 - 2}{6} = \frac{-10}{6} = -\frac{5}{3}$

Verify in the original equation to assure correctness. Hence, both solutions are valid.

Therefore, the solutions are $x_1 = -1$ and $x_2 = -\frac{5}{3}$ , which matches choice 3.

Answer

$x_1=-1,x_2=-\frac{5}{3}$

Exercise #2

Solve the following equation:

$(x-4)^2+3x^2=-16x+12$

Video Solution

Step-by-Step Solution

To solve the given equation, follow these steps:

Step 1: Expand $(x - 4)^2$ using the formula $(a - b)^2 = a^2 - 2ab + b^2$ .

Thus, $(x - 4)^2 = x^2 - 8x + 16$ .

Step 2: Substitute the expanded form into the equation:

$x^2 - 8x + 16 + 3x^2 = -16x + 12$ .

Step 3: Combine like terms on the left-hand side.

This gives $4x^2 - 8x + 16 = -16x + 12$ .

Step 4: Rearrange the equation to set it to zero.

Bring all terms to one side: $4x^2 - 8x + 16 + 16x - 12 = 0$ .

Combine and simplify the terms: $4x^2 + 8x + 4 = 0$ .

Step 5: Simplify the equation by dividing each term by 4.

It becomes $x^2 + 2x + 1 = 0$ .

Step 6: Recognize the equation as a perfect square trinomial.

$(x + 1)^2 = 0$ .

Step 7: Solve by taking the square root of both sides.

The solution is $x + 1 = 0$ , therefore $x = -1$ .

In conclusion, the solution to the equation is $x = -1$ .

Answer

$x=-1$

Exercise #3

Find X

$7x+1+(2x+3)^2=(4x+2)^2$

Video Solution

Step-by-Step Solution

To solve the equation $7x + 1 + (2x + 3)^2 = (4x + 2)^2$ , we follow these steps:

Step 1: Expand both sides using the square of a binomial formula.
Step 2: Simplify the equation to form a standard quadratic equation.
Step 3: Use the quadratic formula to find the roots of the equation.

Step 1: Expand the squares.

The left side: $(2x + 3)^2 = 4x^2 + 12x + 9$ .

The right side: $(4x + 2)^2 = 16x^2 + 16x + 4$ .

Step 2: Substitute back into the original equation and simplify:

$7x + 1 + 4x^2 + 12x + 9 = 16x^2 + 16x + 4$ .

Combine like terms:

$4x^2 + 19x + 10 = 16x^2 + 16x + 4$ .

Step 3: Move all terms to one side:

$4x^2 + 19x + 10 - 16x^2 - 16x - 4 = 0$ .

Which simplifies to:

$-12x^2 + 3x + 6 = 0$ .

Step 4: Divide by -3 to simplify:

$4x^2 - x - 2 = 0$ .

Step 5: Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 4$ , $b = -1$ , $c = -2$ .

Calculate the discriminant:

$b^2 - 4ac = (-1)^2 - 4 \cdot 4 \cdot (-2) = 1 + 32 = 33$ .

Calculate the roots:

$x = \frac{1 \pm \sqrt{33}}{8}$ .

Therefore, the solution to the problem is $x = \frac{1 \pm \sqrt{33}}{8}$ .

Answer

$\frac{1\pm\sqrt{33}}{8}$

Exercise #4

Solve the following equation:

$(x+3)^2=2x+5$

Video Solution

Step-by-Step Solution

To solve the equation $(x+3)^2 = 2x + 5$ , we proceed as follows:

Step 1: Expand the left side. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$ , we find:
$(x+3)^2 = x^2 + 6x + 9$ .
Step 2: Set the equation to zero by moving all terms to one side:
$x^2 + 6x + 9 = 2x + 5$
Subtract $2x + 5$ from both sides:
$x^2 + 6x + 9 - 2x - 5 = 0$
This simplifies to:
$x^2 + 4x + 4 = 0$ .
Step 3: Solve the quadratic equation $x^2 + 4x + 4 = 0$ . Notice this can be factored as:
$(x+2)^2 = 0$ .
Step 4: Solve for $x$ by setting the factor equal to zero:
$x+2 = 0$ .
Thus, $x = -2$ .

Therefore, the solution to the equation $(x+3)^2 = 2x + 5$ is $x = -2$ .

Answer

$x=-2$

Exercise #5

Solve the equation

$2x^2-2x=(x+1)^2$

Video Solution

Step-by-Step Solution

The given equation is:

$2x^2 - 2x = (x+1)^2$

Step 1: Expand the right-hand side.

$(x+1)^2 = x^2 + 2x + 1$

Step 2: Write the full equation with the expanded form.

$2x^2 - 2x = x^2 + 2x + 1$

Step 3: Bring all terms to one side of the equation to set it to zero.

$2x^2 - 2x - x^2 - 2x - 1 = 0$

Step 4: Simplify the equation.

$x^2 - 4x - 1 = 0$

Step 5: Identify coefficients for the quadratic formula.

Here, $a = 1$ , $b = -4$ , $c = -1$ .

Step 6: Apply the quadratic formula.

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}$

$x = \frac{4 \pm \sqrt{16 + 4}}{2}$

$x = \frac{4 \pm \sqrt{20}}{2}$

$x = \frac{4 \pm 2\sqrt{5}}{2}$

$x = 2 \pm \sqrt{5}$

Therefore, the solutions are $x = 2 + \sqrt{5}$ and $x = 2 - \sqrt{5}$ .

These solutions correspond to choice (4): Answers a + b

Answer

Answers a + b

Start practice

Combining the Perfect Square Formulas

Combining the square of a sum formulas

Product of the sum of two terms and their difference

The difference of squares formula

The Square of a Sum Formula

Formulas relating to two expressions cubed

Example

Test yourself on combining short multiplication formulas!

Combining the Shortened Multiplication Formulas

Quadratic Identities

Exercise

Examples with solutions for Combining Short Multiplication Formulas

Exercise #1

Video Solution

Step-by-Step Solution

Answer

Exercise #2

Video Solution

Step-by-Step Solution

Answer

Exercise #3

Video Solution

Step-by-Step Solution

Answer

Exercise #4

Video Solution

Step-by-Step Solution

Answer

Exercise #5

Video Solution

Step-by-Step Solution

Answer