Factoring Trinomials: More than one factorization

True / falseComplete the equationUsing quadrilateralsWorded problemsUsing short multiplication formulasHow many solutions exist to a problem?Solving the equationEquation factorizationTest if the coefficient is different from 1Applying the formulaSolving the problem
Question 1

Solve the following problem:

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

Question 2

Solve the following equation:

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

Question 3

Solve the following problem:

\( (x+1)(x-1)(x+1)=x^2+x^3 \)

Question 4

Solve the following equation:

\( (x-1)(x+1)(x-2)=-2x^2+x^3 \)

Question 5

Solve the following equation:

\( (x+3)^2=(x-3)^2 \)

Examples with solutions for Factoring Trinomials: More than one factorization

Exercise #1

Solve the following problem:

x2+(x−2)2=2(x+1)2 x^2+(x-2)^2=2(x+1)^2

Video Solution

Step-by-Step Solution

Examine the given equation:

x2+(x−2)2=2(x+1)2 x^2+(x-2)^2=2(x+1)^2

Start by simplifying the equation, to achieve this we'll apply the perfect square formula for a binomial squared:

(a±b)2=a2±2ab+b2 (a\pm b)^2=a^2\pm2ab+b^2 ,

Start by opening the parentheses on both sides simultaneously using the perfect square formula, then proceed to combine like terms,

Note that according to the order of operations (which prioritizes exponents over multiplication), the expression inside of the right-hand parentheses is first squared and then the resulting expression is multiplied by 2,

Therefore, the expression that we obtain from applying the perfect square formula on the right-hand side will be placed in parentheses which we'll multiply by 2 (highlighted with an underline in the following calculation):

x2+(x−2)2=2(x+1)2‾↓x2+x2−2⋅x⋅2+22=2(x2+2⋅x⋅1+12)‾x2+x2−4x+4=2(x2+2x+1) x^2+(x-2)^2=2\underline{(x+1)^2} \\ \downarrow\\ x^2+x^2-2\cdot x\cdot2+2^2=2\underline{(x^2+2\cdot x\cdot1+1^2)}\\ x^2+x^2-4x+4=2(x^2+2x+1)\\ Let's continue, open the parentheses on the right side by using the distributive property, move and combine like terms. In the final step we'll solve the simplified equation that we obtain

x2+x2−4x+4=2(x2+2x+1)2x2−4x+4=2x2+4x+2−8x=−2/:(-8)x=−2−8↓x=14 x^2+x^2-4x+4=2(x^2+2x+1)\\ 2x^2-4x+4=2x^2+4x+2\\ -8x=-2\hspace{6pt}\text{/:(-8)}\\ x=\frac{-2}{-8}\\ \downarrow\\ \boxed{x=\frac{1}{4}}

In the final step we reduced the fraction that we obtained as the solution for the unknown,

Therefore the correct answer is answer B.

Answer

x=14 x=\frac{1}{4}

Exercise #2

Solve the following equation:

5x(x+2)(x+5)=5x3 5x(x+2)(x+5)=5x^3

Video Solution

Step-by-Step Solution

Let's examine the given equation:

5x(x+2)(x+5)=5x3 5x(x+2)(x+5)=5x^3

We'll start by opening the second and third pairs of parentheses from the left (marked with an underline below) which are in the left side using the extended distribution law, the result will be placed in new parentheses (since the entire expression is multiplied by the expression that these parentheses multiply) then we'll simplify the expression in the resulting parentheses:

5x(x+2)(x+5)‾=5x3↓5x(x2+2x+5x+10)‾=5x35x(x2+7x+10)‾=5x3 5x\underline{(x+2)(x+5)}=5x^3 \\ \downarrow\\ 5x\underline{\textcolor{blue}{(}x^2+2x+5x+10\textcolor{blue}{)}}=5x^3\\ 5x\underline{\textcolor{blue}{(}x^2+7x+10\textcolor{blue}{)}}=5x^3\\ Continue to use the extended distribution law again and open the parentheses on the left side. Proceed to move and combine like terms:

5x(x2+7x+10)=5x3↓5x3+35x2+50x=5x335x2+50x=0 5x(x^2+7x+10)=5x^3\\ \downarrow\\ 5x^3+35x^2+50x=5x^3\\ 35x^2+50x=0

Note that we obtained a quadratic equation, which can be solved by factoring - by finding a common factor,

Continue to factor out the greatest common factor of the numbers and variables, which is the expression: 5x 5x :

35x2+50x=0↓5x(7x+10)=0 35x^2+50x=0 \\ \downarrow\\ 5x(7x+10)=0

Remember that the product of expressions equals 0 only if at least one of the expressions equals zero, therefore from this equation we obtain two simpler equations:

5x=0/:5x=0 5x=0\hspace{6pt}\text{/}:5\\ \boxed{x=0}

or:

7x+10=07x=−10/:7x=−710 7x+10=0\\ 7x=-10\hspace{6pt}\text{/}:7\\ \boxed{x=-\frac{7}{10}}

Let's summarize the various steps of the solution:

5x(x+2)(x+5)=5x3↓5x(x2+7x+10)=5x3↓5x3+35x2+50x=5x335x2+50x=0↓5x(7x+10)=0↓5x=0→x=07x+10=0→x=−710↓x=0,−710 5x(x+2)(x+5)=5x^3 \\ \downarrow\\ 5x\textcolor{blue}{(}x^2+7x+10\textcolor{blue}{)}=5x^3\\ \downarrow\\ 5x^3+35x^2+50x=5x^3\\ 35x^2+50x=0 \\ \downarrow\\ 5x(7x+10)=0\\ \downarrow\\ 5x=0\rightarrow\boxed{x=0}\\ 7x+10=0\rightarrow\boxed{x=-\frac{7}{10}}\\ \downarrow\\ \boxed{x=0,-\frac{7}{10}}

Therefore the correct answer is answer D.

Answer

A+B are correct.

Exercise #3

Solve the following problem:

(x+1)(x−1)(x+1)=x2+x3 (x+1)(x-1)(x+1)=x^2+x^3

Video Solution

Step-by-Step Solution

Let's examine the given equation:

(x+1)(x−1)(x+1)=x2+x3 (x+1)(x-1)(x+1)=x^2+x^3

Begin by opening the second and third pairs of parentheses from the left (marked with an underline below) On the left side we can apply the difference of squares formula:

(a+b)(a−b)=a2−b2 (a+b)(a-b)=a^2-b^2 ,

Place the result inside of new parentheses (since the resulting expression in its entirety is multiplied by an expression that is enclosed by these parentheses) then we'll proceed to simplify the expression in the resulting parentheses:

(x+1)(x−1)(x+1)‾=x2+x3↓(x+1)(x2−12)‾=x2+x3(x+1)(x2−1)‾=x2+x3 (x+1)\underline{(x-1)(x+1)}=x^2+x^3 \\ \downarrow\\ (x+1)\underline{\textcolor{blue}{(}x^2-1^2\textcolor{blue}{)}}=x^2+x^3\\ (x+1)\underline{\textcolor{blue}{(}x^2-1\textcolor{blue}{)}}=x^2+x^3\\ Continue to use the expanded distributive law again and open the parentheses on the left side. Proceed to move and combine like terms:

(x+1)(x2−1)=x2+x3↓x3−x+x2−1=x2+x3−x=1/⋅(−1)↓x=−1 (x+1)(x^2-1)=x^2+x^3\\ \downarrow\\ x^3-x+x^2-1=x^2+x^3\\ -x =1\hspace{6pt}\text{/}\cdot(-1)\\ \downarrow\\ \boxed{x=-1}

In the final step solve the first-degree equation that we obtained,

Therefore the correct answer is answer A.

Answer

x=−1 x=-1

Exercise #4

Solve the following equation:

(x−1)(x+1)(x−2)=−2x2+x3 (x-1)(x+1)(x-2)=-2x^2+x^3

Video Solution

Step-by-Step Solution

Let's examine the given equation:

(x−1)(x+1)(x−2)=−2x2+x3 (x-1)(x+1)(x-2)=-2x^2+x^3 First, let's simplify the equation, using the perfect square difference formula:

(a+b)(a−b)=a2−b2 (a+b)(a-b)=a^2-b^2 and the expanded distribution law,

We'll start by opening the first pair of parentheses from the left which is in the left side using the perfect square difference formula mentioned, we'll put the result in new parentheses (since the entire expression is multiplied by the expression in the unopened parentheses) then we'll simplify the expression in the parentheses:

(x−1)(x+1)(x−2)=−2x2+x3↓(x2−12)(x−2)=−2x2+x3(x2−1)(x−2)=−2x2+x3 (x-1)(x+1)(x-2)=-2x^2+x^3 \\ \downarrow\\ \textcolor{blue}{(}x^2-1^2\textcolor{blue}{)}(x-2)=-2x^2+x^3\\ \textcolor{blue}{(}x^2-1\textcolor{blue}{)}(x-2)=-2x^2+x^3\\ We'll continue using the expanded distribution law and open the parentheses on the left side, then we'll move terms and combine like terms:

(x2−1)(x−2)=−2x2+x3↓x3−2x2−x+2=−2x2+x3x3−2x2−x+2+2x2−x3=0−x+2=0−x=−2/⋅(−1)x=2 (x^2-1)(x-2)=-2x^2+x^3\\ \downarrow\\ x^3-2x^2-x+2=-2x^2+x^3\\ x^3-2x^2-x+2+2x^2-x^3=0\\ -x+2=0\\ -x=-2\hspace{6pt}\text{/}\cdot(-1)\\ \boxed{x=2}

Let's summarize the equation solving steps:

(x−1)(x+1)(x−2)=−2x2+x3↓(x2−1)(x−2)=−2x2+x3↓x3−2x2−x+2=−2x2+x3x=2 (x-1)(x+1)(x-2)=-2x^2+x^3 \\ \downarrow\\ \textcolor{blue}{(}x^2-1\textcolor{blue}{)}(x-2)=-2x^2+x^3\\ \downarrow\\ x^3-2x^2-x+2=-2x^2+x^3\\ \boxed{x=2} Therefore, the correct answer is answer C.

Answer

x=2 x=2

Exercise #5

Solve the following equation:

(x+3)2=(x−3)2 (x+3)^2=(x-3)^2

Video Solution

Step-by-Step Solution

Let's examine the given equation:

(x+3)2=(x−3)2 (x+3)^2=(x-3)^2 First, let's simplify the equation, for this we'll use the perfect square formula for a binomial squared:

(a±b)2=a2±2ab+b2 (a\pm b)^2=a^2\pm2ab+b^2 ,

We'll start by opening the parentheses on both sides simultaneously using the perfect square formula mentioned, then we'll move terms and combine like terms, and in the final step we'll solve the resulting simplified equation:

(x+3)2=(x−3)2↓x2+2⋅x⋅3+32=x2−2⋅x⋅3+32x2+6x+9=x2−6x+9x2+6x+9−x2+6x−9=012x=0/:12x=0 (x+3)^2=(x-3)^2 \\ \downarrow\\ x^2+2\cdot x\cdot3+3^2= x^2-2\cdot x\cdot3+3^2 \\ x^2+6x+9= x^2-6x+9 \\ x^2+6x+9- x^2+6x-9 =0\\ 12x=0\hspace{6pt}\text{/}:12\\ \boxed{x=0} Therefore, the correct answer is answer A.

Answer

x=0 x=0

Question 1

Solve the following equation:

\( (x+3)(x-3)=x^2+x \)

Question 2

\( (x-4)^2=(x+2)(x-1) \)

Exercise #6

Solve the following equation:

(x+3)(x−3)=x2+x (x+3)(x-3)=x^2+x

Video Solution

Step-by-Step Solution

Let's examine the given equation:

(x+3)(x−3)=x2+x (x+3)(x-3)=x^2+x First, let's simplify the equation, for this we'll use the difference of squares factoring formula:

(a+b)(a−b)=a2−b2 (a+b)(a-b)=a^2-b^2 ,

We'll start by opening the parentheses on the left side using the mentioned factoring formula, then we'll move terms, combine like terms, and finally solve the resulting equation:

(x+3)(x−3)=x2+x↓x2−32=x2+xx2−9=x2+xx2−9−x2−x=0x=−9 (x+3)(x-3)=x^2+x \\ \downarrow\\ x^2-3^2=x^2+x \\ x^2-9=x^2+x \\ x^2-9-x^2-x=0 \\ \boxed{x=-9} Therefore, the correct answer is answer B.

Answer

x=−9 x=-9

Exercise #7

(x−4)2=(x+2)(x−1) (x-4)^2=(x+2)(x-1)

Video Solution

Step-by-Step Solution

Let's examine the given equation:

(x−4)2=(x+2)(x−1) (x-4)^2=(x+2)(x-1) First, let's simplify the equation, using the perfect square binomial formula:

(a±b)2=a2±2ab+b2 (a\pm b)^2=a^2\pm2ab+b^2 and the expanded distributive law,

We'll start by opening the parentheses using the perfect square binomial formula mentioned and using the expanded distributive law and then we'll move terms and combine like terms:

(x−4)2=(x+2)(x−1)↓x2−2⋅x⋅4+42=x2−x+2x−2x2−8x+16=x2−x+2x−2x2−8x+16−x2+x−2x+2=0−9x+18=0 (x-4)^2=(x+2)(x-1) \\ \downarrow\\ x^2-2\cdot x\cdot4+4^2=x^2-x+2x-2\\ x^2-8x+16=x^2-x+2x-2\\ x^2-8x+16-x^2+x-2x+2=0\\ -9x+18=0 We got a first-degree equation, we'll solve it in the regular way by isolating the variable on one side:

−9x+18=0−9x=−18/:(−9)x=2 -9x+18=0 \\ -9x=-18\hspace{6pt}\text{/}:(-9)\\ \boxed{x=2}

Let's summarize the equation solving steps:

(x−4)2=(x+2)(x−1)↓x2−8x+16=x2−x+2x−2−9x+18=0↓x=2 (x-4)^2=(x+2)(x-1) \\ \downarrow\\ x^2-8x+16=x^2-x+2x-2\\ -9x+18=0 \\ \downarrow\\ \boxed{x=2} Therefore, the correct answer is answer C.

Answer

x=2 x=2