Factoring Trinomials: True / false

Complete the equationUsing quadrilateralsWorded problemsUsing short multiplication formulasHow many solutions exist to a problem?More than one factorizationSolving the equationEquation factorizationTest if the coefficient is different from 1Applying the formulaSolving the problem
Question 1

The decomposition of the previous trinomial:

\( x^2+16x+64=0 \)

is \( (x+8)(x+8)=0 \)

Question 2

The decomposition of the previous trinomial:

\( x^2+3x-18=0 \)

is \( (x+6)(x-3)=0 \)

Question 3

The decomposition of the previous trinomial:

\( x^2-5x+6=0 \)

is \( (x-5)(x+6) \)

Question 4

The decomposition of the previous trinomial:

\( x^2+x-20=0 \)

is \( (x+2)(x-10)=0 \)

Examples with solutions for Factoring Trinomials: True / false

Exercise #1

The decomposition of the previous trinomial:

x2+16x+64=0 x^2+16x+64=0

is (x+8)(x+8)=0 (x+8)(x+8)=0

Video Solution

Step-by-Step Solution

Let's try to factor using quick trinomial factoring the given expression:

x2+16x+64 x^2+16x+64\\

Let's look for a pair of numbers whose product is the free term in the expression, and their sum is the coefficient of the first-degree term in the expression, meaning two numbers m,n m,\hspace{2pt}n that satisfy:

m⋅n=64m+n=16 m\cdot n=64\\ m+n=16\\ From the first requirement mentioned, namely- from the multiplication, note that the product of the numbers we're looking for must yield a positive result and therefore we can conclude that both numbers have equal signs, according to multiplication rules, now we'll remember that 64 has several possible pairs of whole number factors, we won't list all possibilities here, but note that:

64=82=8⋅8 64=8^2=8\cdot8 and- 16=8+8 16=8+8 Continuing, meeting the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are equal to each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

{m=8n=8 \begin{cases} m=8\\ n=8\end{cases}

(We can check all other factor pairs of 64 to verify this is the only possibility, but once a suitable option is found - it must be the only one)

Therefore we can factor the given expression to:

x2+16x+64↓(x+8)(x+8) x^2+16x+64\\ \downarrow\\ (x+8)(x+8)

And therefore clearly the suggested factorization in the problem is correct.

That is - the correct answer is answer A.

Note:

The given question could also be solved by expanding the parentheses in the suggested expression:

(x+8)(x+8) (x+8)(x+8) (using the extended distribution law or alternatively using the shortened multiplication formula for squared binomial in this case), and checking if indeed we get the given expression:

x2+16x+64 x^2+16x+64 , however it's obviously better to try to factor the given expression- for practice purposes.

Answer

True

Exercise #2

The decomposition of the previous trinomial:

x2+3x−18=0 x^2+3x-18=0

is (x+6)(x−3)=0 (x+6)(x-3)=0

Video Solution

Step-by-Step Solution

Let's try to factorize the given expression using quick trinomial factoring:

x2+3x−18 x^2+3x-18

We'll look for a pair of numbers whose product is the free term in the expression, and whose sum is the coefficient of the first-degree term, meaning two numbers m,n m,\hspace{2pt}n that satisfy:

m⋅n=−18m+n=3 m\cdot n=-18\\ m+n=3\\ From the first requirement mentioned, that is - from the multiplication, we should note that the product of the numbers we're looking for needs to yield a negative result, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 18 are 2 and 9, 6 and 3, or 18 and 1. Meeting the second requirement mentioned, along with the fact that the numbers we're looking for have different signs leads to the conclusion that the only possibility for the two numbers we're looking for is:

{m=6n=−3 \begin{cases} m=6\\ n=-3 \end{cases}

Therefore we'll factorize the given expression to:

x2+3x−18↓(x+6)(x−3) x^2+3x-18 \\ \downarrow\\ (x+6)(x-3)

Thus clearly the suggested factorization in the problem is correct.

Therefore - the correct answer is answer A.

Note:

The given question could also be solved by expanding the parentheses in the suggested expression:

(x+6)(x−3) (x+6)(x-3) (using the expanded distributive law), and checking if indeed we get the given expression:

x2+3x−18 x^2+3x-18 , however it is of course preferable to try to factorize the given expression - for practice purposes.

Answer

True

Exercise #3

The decomposition of the previous trinomial:

x2−5x+6=0 x^2-5x+6=0

is (x−5)(x+6) (x-5)(x+6)

Video Solution

Step-by-Step Solution

Let's try to factor using quick trinomial factoring the given expression:

x2−5x+6 x^2-5x+6

We'll look for a pair of numbers whose product is the free term in the expression, and their sum is the coefficient of the first power term in the expression, meaning two numbers m,n m,\hspace{2pt}n that satisfy:

m⋅n=6m+n=−5 m\cdot n=6\\ m+n=-5\\ From the first requirement mentioned, namely- from the multiplication, we should note that the product of the numbers we're looking for needs to yield a positive result and therefore we can conclude that both numbers have the same signs, according to multiplication rules, and now we'll remember that the possible factors of 6 are 2 and 3 or 6 and 1, fulfilling the second requirement mentioned, together with the fact that the signs of the numbers we're looking for are equal to each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

{m=−3n=−2 \begin{cases} m=-3\\ n=-2 \end{cases}

and therefore we'll factor the given expression to:

x2−5x+6↓(x−3)(x−2) x^2-5x+6 \\ \downarrow\\ (x-3)(x-2)

and therefore clearly the factorization suggested in the problem is incorrect.

Therefore- the correct answer is answer B.

Note:

The given question could also be solved by expanding the parentheses in the suggested expression:

(x−5)(x+6) (x-5)(x+6) (using the expanded distributive property), and checking if indeed we get the given expression:

x2−5x+6 x^2-5x+6 , however it is of course preferable to try to factor the given expression- for practice purposes.

Answer

Not true

Exercise #4

The decomposition of the previous trinomial:

x2+x−20=0 x^2+x-20=0

is (x+2)(x−10)=0 (x+2)(x-10)=0

Video Solution

Step-by-Step Solution

Let's try to factorize the given expression using quick trinomial factoring:

x2+x−20 x^2+x-20

We'll look for a pair of numbers whose product is the free term in the expression, and whose sum is the coefficient of the first power term in the expression, meaning two numbers m,n m,\hspace{2pt}n that satisfy:

m⋅n=−20m+n=1 m\cdot n=-20\\ m+n=1\\ From the first requirement mentioned, namely - from the multiplication, we should note that the product of the numbers we're looking for needs to yield a negative result, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 20 are 2 and 10, 4 and 5, or 20 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are different from each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

{m=5n=−4 \begin{cases} m=5\\ n=-4 \end{cases}

and therefore we'll factorize the given expression to:

x2+x−20↓(x+5)(x−4) x^2+x-20 \\ \downarrow\\ (x+5)(x-4)

and therefore clearly the suggested factorization in the problem is incorrect.

Therefore - the correct answer is answer B.

Note:

The given question could also be solved by expanding the parentheses in the suggested expression:

(x+2)(x−10) (x+2)(x-10) (using the expanded distributive property), and checking if it indeed equals the given expression:

x2+x−20 x^2+x-20 , however it is of course preferable to try to factorize the given expression - for practice purposes.

Answer

Not true