Question

factor by grouping xv-5x-2v+10

Answer 1

ok so what is the first step

Answer 2

*) Factor 2 x 3 + 10 x 2 + 3 x + 15 2x^3+10x^2+3x+15 2x3+10x2+3x+152, x, cubed, plus, 10, x, squared, plus, 3, x, plus, 15.

Answer 3

do you have any answer choices?

Answer 4

no comprende

Answer 5

*) Factor 2 x 3 + 10 x 2 + 3 x + 15 2x^3+10x^2+3x+15 2x3+10x2+3x+152, x, cubed, plus, 10, x, squared, plus, 3, x, plus, 15.

Answer 6

no i dont have any answer choices

Answer 7

ok

Answer 8

or maybe a screen shot

Answer 9

just put this
*) Factor 2 x 3 + 10 x 2 + 3 x + 15 2x^3+10x^2+3x+15 2x3+10x2+3x+152, x, cubed, plus, 10, x, squared, plus, 3, x, plus, 15.

Answer 10

put it in answer form coco

Answer 11

that is the answer

Answer 12

Example 1: Factoring 2x^2+8x+3x+122x
2
+8x+3x+122, x, squared, plus, 8, x, plus, 3, x, plus, 12
First, notice that there is no factor common to all terms in 2x^2+8x+3x+122x
2
+8x+3x+122, x, squared, plus, 8, x, plus, 3, x, plus, 12. However, if we group the first two terms together and the last two terms together, each group has its own GCF, or greatest common factor:


In particular, there is a GCF of 2x2x2, x in the first grouping and a GCF of 333 in the second grouping. We can factor these out to obtain the following expression:
2x(x+4)+3(x+4)2x(x+4)+3(x+4)2, x, left parenthesis, x, plus, 4, right parenthesis, plus, 3, left parenthesis, x, plus, 4, right parenthesis
[How did you do that?]
Notice that this reveals yet another common factor between the two terms: \goldD{x+4}x+4start color #e07d10, x, plus, 4, end color #e07d10. We can use the distributive property to factor out this common factor.

Since the polynomial is now expressed as a product of two binomials, it is in factored form. We can check our work by multiplying and comparing it to the original polynomial. [I'd like to see this, please!]
Example 2: Factoring 3x^2+6x+4x+83x
2
+6x+4x+83, x, squared, plus, 6, x, plus, 4, x, plus, 8
Let's summarize what was done above by factoring another polynomial.
\begin{aligned}&\phantom{=}3x^2+6x+4x+8\\\\ &=(3x^2+6x)+(4x+8)&&\small{\gray{\text{Group terms}}}\\ \\ &=3x({x+2})+4({x+2})&&\small{\gray{\text{Factor out GCFs}}}\\ \\ &=3x(\goldD{x+2})+4(\goldD{x+2})&&\small{\gray{\text{Common factor!}}}\\\\ &=(\goldD{x+2})(3x+4)&&\small{\gray{\text{Factor out } x+2}} \end{aligned}
​

=3x
2
+6x+4x+8
=(3x
2
+6x)+(4x+8)
=3x(x+2)+4(x+2)
=3x(x+2)+4(x+2)
=(x+2)(3x+4)
​

​

Group terms
Factor out GCFs
Common factor!
Factor out x+2
​

The factored form is (x+2)(3x+4)(x+2)(3x+4)left parenthesis, x, plus, 2, right parenthesis, left parenthesis, 3, x, plus, 4, right parenthesis.
Check your understanding
1) Factor 9x^2+6x+12x+89x
2
+6x+12x+89, x, squared, plus, 6, x, plus, 12, x, plus, 8.
Choose 1 answer:
Choose 1 answer:

(Choice A)
A
(3x+2)(3x+4)(3x+2)(3x+4)left parenthesis, 3, x, plus, 2, right parenthesis, left parenthesis, 3, x, plus, 4, right parenthesis

(Choice B)
B
(3x+1)(3x+8)(3x+1)(3x+8)left parenthesis, 3, x, plus, 1, right parenthesis, left parenthesis, 3, x, plus, 8, right parenthesis

(Choice C)
C
(9x+6)(12x+8)(9x+6)(12x+8)left parenthesis, 9, x, plus, 6, right parenthesis, left parenthesis, 12, x, plus, 8, right parenthesis

(Choice D)
D
(3x+2)(3x+2)(3x+4)(3x+2)(3x+2)(3x+4)left parenthesis, 3, x, plus, 2, right parenthesis, left parenthesis, 3, x, plus, 2, right parenthesis, left parenthesis, 3, x, plus, 4, right parenthesis
[I need help!]
2) Factor 5x^2+10x+2x+45x
2
+10x+2x+45, x, squared, plus, 10, x, plus, 2, x, plus, 4.
[I need help!]
3) Factor 8x^2+6x+4x+38x
2
+6x+4x+38, x, squared, plus, 6, x, plus, 4, x, plus, 3.
[I need help!]
Example 3: Factoring 3x^2-6x-4x+83x
2
−6x−4x+83, x, squared, minus, 6, x, minus, 4, x, plus, 8
Extra care should be taken when using the grouping method to factor a polynomial with negative coefficients.
For example, the steps below can be used to factor 3x^2-6x-4x+83x
2
−6x−4x+83, x, squared, minus, 6, x, minus, 4, x, plus, 8.
\begin{aligned}\phantom{0}&&&\phantom{=}3x^2-6x-4x+8\\\\ \small{\blueD{(1)}}&&&=(3x^2-6x)+(-4x+8)&&\small{\gray{\text{Group terms}}}\\\\ \small{\blueD{(2)}}&&&=3x(x-2)+(-4)(x-2)&&\small{\gray{\text{Factor out GCFs}}}\\\\ \small{\blueD{(3)}}&&&=3x(x-2)-4(x-2)&&\small{\gray{\text{Simplify}}}\\\\ \small{\blueD{(4)}}&&&=3x(\goldD{x-2})-4(\goldD{x-2})&&\small{\gray{\text{Common factor!}}}\\\\ \small{\blueD{(5)}}&&&=(\goldD{x-2})(3x-4)&&\small{\gray{\text{Factor out $x-2$}}}\\\\ \end{aligned}
0
(1)
(2)
(3)
(4)
(5)
​

​

​

=3x
2
−6x−4x+8
=(3x
2
−6x)+(−4x+8)
=3x(x−2)+(−4)(x−2)
=3x(x−2)−4(x−2)
=3x(x−2)−4(x−2)
=(x−2)(3x−4)
​

​

Group terms
Factor out GCFs
Simplify
Common factor!
Factor out x−2
​

The factored form of the polynomial is (x-2)(3x-4)(x−2)(3x−4)left parenthesis, x, minus, 2, right parenthesis, left parenthesis, 3, x, minus, 4, right parenthesis. We can multiply the binomials to check our work. [I'd like to see this, please!]
A few of the steps above may seem different than what you saw in the first example, so you may have a few questions.
Where did the "+" sign between the groupings come from?
In step \small{\blueD{(1)}}(1)start color #11accd, left parenthesis, 1, right parenthesis, end color #11accd, a "+" sign was added between the groupings (3x^2-6x)(3x
2
−6x)left parenthesis, 3, x, squared, minus, 6, x, right parenthesis and (-4x+8)(−4x+8)left parenthesis, minus, 4, x, plus, 8, right parenthesis. This is because the third term (-4x)(−4x)left parenthesis, minus, 4, x, right parenthesis is negative, and the sign of the term must be included within the grouping.
Keeping the minus sign outside the second grouping is tricky. For example, a common error is to group 3x^2-6x-4x+83x
2
−6x−4x+83, x, squared, minus, 6, x, minus, 4, x, plus, 8 as (3x^2-6x)-(4x+8)(3x
2
−6x)−(4x+8)left parenthesis, 3, x, squared, minus, 6, x, right parenthesis, minus, left parenthesis, 4, x, plus, 8, right parenthesis. This grouping, however, simplifies to 3x^2-6x-4x\maroonD{-8}3x
2
−6x−4x−83, x, squared, minus, 6, x, minus, 4, x, start color #ca337c, minus, 8, end color #ca337c, which is not the same as the original expression. [Can I see this?]
Why factor out -4−4minus, 4 instead of 444?
In step \small{\blueD{(2)}}(2)start color #11accd, left parenthesis, 2, right parenthesis, end color #11accd, we factored out a -4−4minus, 4 to reveal a common factor of (x-2)(x−2)left parenthesis, x, minus, 2, right parenthesis between the terms. If we instead factored out a positive 444, we would not obtain that common binomial factor seen above:
\begin{aligned}(3x^2-6x)+(-4x+8)&=3x(\goldD{x-2})+4(\purpleC{-x+2})\\ \end{aligned}
(3x
2
−6x)+(−4x+8)
​

=3x(x−2)+4(−x+2)
​

When the leading term in a group is negative, we will often need to factor out a negative common factor.
Check your understanding
4) Factor 2x^2-3x-4x+62x
2
−3x−4x+62, x, squared, minus, 3, x, minus, 4, x, plus, 6.
Choose 1 answer:
Choose 1 answer:

(Choice A)
A
(2x+3)(x-2)(2x+3)(x−2)left parenthesis, 2, x, plus, 3, right parenthesis, left parenthesis, x, minus, 2, right parenthesis

(Choice B)
B
(2x-3)(x-2)(2x−3)(x−2)left parenthesis, 2, x, minus, 3, right parenthesis, left parenthesis, x, minus, 2, right parenthesis

(Choice C)
C
(2x+3)(2x-3)(x-2)(2x+3)(2x−3)(x−2)left parenthesis, 2, x, plus, 3, right parenthesis, left parenthesis, 2, x, minus, 3, right parenthesis, left parenthesis, x, minus, 2, right parenthesis
[I need help!]
5) Factor 3x^2+3x-10x-103x
2
+3x−10x−103, x, squared, plus, 3, x, minus, 10, x, minus, 10.
[I need help!]
6) Factor 3x^2+6x-x-23x
2
+6x−x−23, x, squared, plus, 6, x, minus, x, minus, 2.
[I need help!]
Challenge problem
7*) Factor 2x^3+10x^2+3x+152x
3
+10x
2
+3x+152, x, cubed, plus, 10, x, squared, plus, 3, x, plus, 15.
[I need help!]
When can we use the grouping method?
The grouping method can be used to factor polynomials whenever a common factor exists between the groupings.
For example, we can use the grouping method to factor 3x^2+9x+2x+63x
2
+9x+2x+63, x, squared, plus, 9, x, plus, 2, x, plus, 6 since it can be written as follows:
\begin{aligned}(3x^2+9x)+(2x+6)&=3x(\goldD{x+3})+2(\goldD{x+3})\\ \end{aligned}
(3x
2
+9x)+(2x+6)
​

=3x(x+3)+2(x+3)
​

We cannot, however, use the grouping method to factor 2x^2+3x+4x+122x
2
+3x+4x+122, x, squared, plus, 3, x, plus, 4, x, plus, 12 because factoring out the GCF from both groupings does not yield a common factor!
\begin{aligned}(2x^2+3x)+(4x+12)&=x(\goldD{2x+3})+4(\purpleC{x+3})\\ \end{aligned}
(2x
2
+3x)+(4x+12)
​

=x(2x+3)+4(x+3)
​

Using grouping to factor trinomials
You can also use grouping to factor certain three termed quadratics (i.e. trinomials) like 2x^2+7x+32x
2
+7x+32, x, squared, plus, 7, x, plus, 3. This is because we can rewrite the expression as follows:
2x^2+\blueD7x+3=2x^2+\blueD1x+\blueD6x+32x
2
+7x+3=2x
2
+1x+6x+32, x, squared, plus, start color #11accd, 7, end color #11accd, x, plus, 3, equals, 2, x, squared, plus, start color #11accd, 1, end color #11accd, x, plus, start color #11accd, 6, end color #11accd, x, plus, 3
Then we can use grouping to factor 2x^2+\blueD1x+\blueD6x+32x
2
+1x+6x+32, x, squared, plus, start color #11accd, 1, end color #11accd, x, plus, start color #11accd, 6, end color #11accd, x, plus, 3 as (x+3)(2x+1)(x+3)(2x+1)left parenthesis, x, plus, 3, right parenthesis, left parenthesis, 2, x, plus, 1, right parenthesis.
For more on factoring quadratic trinomials like these using the grouping method, check out our next article.
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Nin
4 years ago
Posted 4 years ago. Direct link to Nin's post “This may sound a bit dumb, but is there any signif...”
This may sound a bit dumb, but is there any significant difference between factoring, grouping trinomials, difference of squares, and GCF?

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Copperhead514
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Posted a year ago. Direct link to Copperhead514's post “Nothing significant, but there are important (howe...”
Nothing significant, but there are important (however small) differences and they are used for different things.

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daleighellison
3 years ago
Posted 3 years ago. Direct link to daleighellison's post “How would you work out the problem if there were o...”
How would you work out the problem if there were only 3 terms?

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Ian Pulizzotto
3 years ago
Posted 3 years ago. Direct link to Ian Pulizzotto's post “Find two numbers that add to the middle coefficien...”
Great Answer
Good Answer
Find two numbers that add to the middle coefficient and multiply to give the product of the first and last coefficients (or constants). This is called the ac method.

Example: Factor 6x^2 + 19x + 10.
6*10 = 60, so we need to find two numbers that add to 19 and multiply to give 60. These numbers (after some trial and error) are 15 and 4. So split up 19x into 15x + 4x (or 4x + 15x), then factor by grouping:

6x^2 + 19x + 10 = 6x^2 + 15x + 4x + 10
= 3x(2x + 5) + 2(2x + 5)
= (3x + 2)(2x + 5).

Have a blessed, wonderful day!

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rama
3 years ago
Posted 3 years ago. Direct link to rama's post “what the meaning of GCF?”
what the meaning of GCF?

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Kim Seidel
3 years ago
Posted 3 years ago. Direct link to Kim Seidel's post “GCF is the abbreviation for Greatest Common Factor...”
GCF is the abbreviation for Greatest Common Factor.
It is the value that you can evenly divide all terms by. It can be a number, a variable, or a mix of numbers and variables.

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Hafsa Mohamed
3 years ago
Posted 3 years ago. Direct link to Hafsa Mohamed's post “In problem 3, I solved the expression and got the ...”
In problem 3, I solved the expression and got the answer (4x+2)(2x+1.5). It said my answer was correct, but when I checked my answer, I got the same expression. What did I do wrong?

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Nikkia Metsker
9 months ago
Posted 9 months ago. Direct link to Nikkia Metsker's post “What is the point of this? I'm not going to use it...”
What is the point of this? I'm not going to use it when i do daily life things.

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J E
9 months ago
Posted 9 months ago. Direct link to J E's post “You might be surprised! If you decide to go into a...”
You might be surprised! If you decide to go into a career like engineering or to work on cool projects like an autonomous car or a rocket, you will use polynomials on a daily basis to solve problems like how to control the car using different sensor inputs. Knowing how to manipulate polynomials like this is therefore extremely useful when designing these systems!

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katelynn Rodriguez
2 years ago
Posted 2 years ago. Direct link to katelynn Rodriguez's post “so for your polynomials that you can not use the g...”
so for your polynomials that you can not use the grouping method for what do you do to solve them ?

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David Lee
2 years ago
Posted 2 years ago. Direct link to David Lee's post “There is a formula for that. Here is the link (for...”
There is a formula for that. Here is the link (for quadratic): https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/a/quadratic-formula-review

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2 years ago
Posted 2 years ago. Direct link to Lavont's post “How can I use the distributive property twice?”
How can I use the distributive property twice?

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Aaishahali
3 years ago
Posted 3 years ago. Direct link to Aaishahali's post “can someone help me with these type of questions F...”
can someone help me with these type of questions
Factor as the product of two binomials.
x​​ +8x+12

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David Matisko
3 years ago
Posted 3 years ago. Direct link to David Matisko's post “Is it supposed to be: X^2 +8x +12 ? I just noticed...”
Is it supposed to be: X^2 +8x +12 ?
I just noticed that the power of 2 is missing.
The first step is the hardest step. Look at the 12 and break it down to pairs of factors. The pairs are 1*12, 2*6, or 3*4. Now find the pair that adds to 8; That is 2*6! okay that was the hard part, now to finish it:
x^2+8x+12 first step, change the 8 to a 2&6:
x^2+2x+6x+12 see how this is equivalent? Now factor by grouping.
x(x+2)+6(x+2) got the same factor! (x+2)
(x+2)(x+6) all done. (If we FOIL this we get back to the start.)
Note: if there is at least one negative sign we might have to subtract instead of adding.


Here is one with a negative sign ( a little harder ):
x^2+10x-24
The pairs of factors that make 24 are 1*24, 2*12, 3*8, or 4*6. Because the 24 is negative we need the pair that subtracts to 10; Which is 2*12 (not 4*6)
X^2-2x+12x-24 (note that we need a negative 2 and positive 12 to make 10) Now factor by grouping
x(x-2)+12(x-2) Now factor the polynomial with a common binomial
(x-2)(x+12) (FOIL this to get back to the start)

Note: Need to know (factor by grouping) & (factor polynomials with a common binomial)
before tackling this kind of problem
I found a video on this called: Factoring quadratics by grouping

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Dan Madsen
3 years ago
Posted 3 years ago. Direct link to Dan Madsen's post “How does one factor by grouping when their is more...”
How does one factor by grouping when their is more than one letter in the equation. For example: 8ax+18a+20x+45

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David Severin
3 years ago
Posted 3 years ago. Direct link to David Severin's post “well, if you factor the first two to get 2a (4x + ...”
well, if you factor the first two to get 2a (4x + 9) and the second two as 5 (4x+9), you can see that an equivalent would be (2a + 5)(4x+9). Or if you factored first and third tern and second and fourth, you get 4x(2a+5) + 9(2a+5) and get the same answer

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Ezra
2 years ago
Posted 2 years ago. Direct link to Ezra 's post “How can I factor by grouping this expression: 2m^2...”
How can I factor by grouping this expression: 2m^2 - 3t - 6m + mt
Is it possible?

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Kim Seidel
2 years ago
Posted 2 years ago. Direct link to Kim Seidel's post “Reorder the terms: 2m^2 - 6m + mt - 3t Then, use g...”
Reorder the terms: 2m^2 - 6m + mt - 3t
Then, use grouping.
Factor out 2m from 1st pair and t from 2nd pair
2m(m-3) + t(m-3)
Factor out (m-3)
(m-3)(2m+t)

Hope this helps.

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Intro to groupingFactoring quadratics by

Answer 13

Equation at the end of step 1 (((2 • (x3)) + (2•5x2)) - 3x) - 15 STEP 2: ((2x3 + (2•5x2)) - 3x) - 15 STEP3:Checking for a perfect cube 3.1 2x3+10x2-3x-15 is not a perfect cube Trying to factor by pulling out : 3.2 Factoring: 2x3+10x2-3x-15 Thoughtfully split the expression at hand into groups, each group having two terms : Group 1: -3x-15 Group 2: 2x3+10x2 Pull out from each group separately : Group 1: (x+5) • (-3) Group 2: (x+5) • (2x2) ------------------- Add up the two groups : (x+5) • (2x2-3) Which is the desired factorization Trying to factor as a Difference of Squares: 3.3 Factoring: 2x2-3 Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B) Proof : (A+B) • (A-B) = A2 - AB + BA - B2 = A2 - AB + AB - B2 = A2 - B2 Note : AB = BA is the commutative property of multiplication. Note : - AB + AB equals zero and is therefore eliminated from the expression. Check : 2 is not a square !! Ruling : Binomial can not be factored as the difference of two perfect squares Final result : (2x2 - 3) • (x + 5)

Answer 14

shorten it
(2x2 - 3) • (x + 5)