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x + a x a = x 2 a 2 size 12{ left (x+a right ) left (x - a right )=x rSup { size 8{2} } - a rSup { size 8{2} } } {}

You can run this formula in reverse whenever you are subtracting two perfect squares . For instance, if we see x 2 25 size 12{x rSup { size 8{2} } - "25"} {} , we recognize that both x 2 size 12{x rSup { size 8{2} } } {} and 25 are perfect squares. We can therefore factor it as x + 5 x 5 size 12{ left (x+5 right ) left (x - 5 right )} {} . Other examples include:

  • x 2 64 size 12{x rSup { size 8{2} } - "64"} {} = x + 8 x 8 size 12{ {}= left (x+8 right ) left (x - 8 right )} {}
  • 16 y 2 49 size 12{"16"y rSup { size 8{2} } - "49"} {} = 4y + 7 4y 7 size 12{ {}= left (4y+7 right ) left (4y - 7 right )} {}
  • 2x 2 18 size 12{2x rSup { size 8{2} } - "18"} {} = 2 x 2 9 size 12{ {}=2 left (x rSup { size 8{2} } - 9 right )} {} = 2 x + 3 x 3 size 12{ {}=2 left (x+3 right ) left (x - 3 right )} {}

And so on. Note that, in the last example, we begin by pulling out a 2, and we are then left with two perfect squares. This is an example of the rule that you should always begin by pulling out common factors before you try anything else!

It is also important to note that you cannot factor the sum of two squares. x 2 + 4 size 12{x rSup { size 8{2} } +4} {} is a perfectly good function, but it cannot be factored.

Brute force, old-fashioned, bare-knuckle, no-holds-barred factoring

In this case, the multiplication that we are reversing is just FOIL. For instance, consider:

x + 3 x + 7 = x 2 + 3x + 7x + 21 = x 2 + 10 x + 21 size 12{ left (x+3 right ) left (x+7 right )=x rSup { size 8{2} } +3x+7x+"21"=x rSup { size 8{2} } +"10"x+"21"} {}

What happened? The 3 and 7 added to yield the middle term (10), and multiplied to yield the final term 21 size 12{ left ("21" right )} {} . We can generalize this as: x + a x + b = x 2 + a + b x + ab size 12{ left (x+a right ) left (x+b right )=x rSup { size 8{2} } + left (a+b right )x+ ital "ab"} {} .

The point is, if you are given a problem such as x 2 + 10 x + 21 size 12{x rSup { size 8{2} } +"10"x+"21"} {} to factor, you look for two numbers that add up to 10, and multiply to 21. And how do you find them? There are a lot of pairs of numbers that add up to 10, but relatively few that multiply to 21. So you start by looking for factors of 21.

Below is a series of examples. Each example showcases a different aspect of the factoring process, so I would encourage you not to skip over any of them: try each problem yourself, then take a look at what I did.

If you are uncomfortable with factoring, the best practice you can get is to multiply things out . In each case, look at the final answer I arrive at, and multiply it with FOIL. See that you get the problem I started with. Then look back at the steps I took and see how they led me to that answer. The steps will make a lot more sense if you have done the multiplication already.

Factor x 2 + 11 x + 18 size 12{x rSup { size 8{2} } +"11"x+"18"} {}

x + __ x + __ size 12{ left (x+"__" right ) left (x+"__" right )} {}

What multiplies to 18? 1 18 size 12{1 cdot "18"} {} , or 2 9 size 12{2 cdot 9} {} , or 3 6 size 12{3 cdot 6} {} .

Which of those adds to 11? 2 + 9 size 12{2+9} {} .

x + 2 x + 9 size 12{ left (x+2 right ) left (x+9 right )} {}

Start by listing all factors of the third term. Then see which ones add to give you the middle term you want.

Factor x 2 13 x + 12 size 12{x rSup { size 8{2} } - "13"x+"12"} {}

x + __ x + __ size 12{ left (x+"__" right ) left (x+"__" right )} {}

What multiplies to 12? 1 12 size 12{1 cdot "12"} {} , or 2 6 size 12{2 cdot 6} {} , or 3 4 size 12{3 cdot 4} {}

Which of those adds to 13? 1 + 12 size 12{1+"12"} {}

x 1 x 12 size 12{ left (x - 1 right ) left (x - "12" right )} {}

If the middle term is negative, it doesn’t change much: it just makes both numbers negative. If this had been x 2 + 13 x + 12 size 12{x rSup { size 8{2} } +"13"x+"12"} {} , the process would have been the same, and the answer would have been x + 1 x + 12 size 12{ left (x+1 right ) left (x+"12" right )} {} .

Factor x 2 + 12 x + 24 size 12{x rSup { size 8{2} } +"12"x+"24"} {}

x + __ x + __ size 12{ left (x+"___" right ) left (x+"___" right )} {}

What multiplies to 24? 1 24 size 12{1 cdot "24"} {} , or 2 12 size 12{2 cdot "12"} {} , or 3 8 size 12{3 cdot 8} {} , or 4 6 size 12{4 cdot 6} {}

Which of those adds to 12? None of them.

It can’t be factored. It is “prime.”

Some things can’t be factored. Many students spend a long time fighting with such problems, but it really doesn’t have to take long. Try all the possibilities, and if none of them works, it can’t be factored.

Factor x 2 + 2x 15 size 12{x rSup { size 8{2} } +2x - "15"} {}

x + __ x + __ size 12{ left (x+"___" right ) left (x+"___" right )} {}

What multiplies to 15? 1 15 size 12{1 cdot "15"} {} , or 3 5 size 12{3 cdot 5} {}

Which of those subtracts to 2? 5–3

x + 5 x 3 size 12{ left (x+5 right ) left (x - 3 right )} {}

If the last term is negative, that changes things! In order to multiply to –15, the two numbers will have to have different signs—one negative, one positive—which means they will subtract to give the middle term. Note that if the middle term were negative, that wouldn’t change the process: the final answer would be reversed, x + 5 x 3 size 12{ left (x+5 right ) left (x - 3 right )} {} . This fits the rule that we saw earlier—changing the sign of the middle term changes the answer a bit, but not the process.

Factor 2x 2 + 24 x + 72 size 12{2x rSup { size 8{2} } +"24"x+"72"} {}

2 x 2 + 12 x + 36 size 12{2 left (x rSup { size 8{2} } +"12"x+"36" right )} {}

2 x + 6 2 size 12{2 left (x+6 right ) rSup { size 8{2} } } {}

Never forget, always start by looking for common factors to pull out. Then look to see if it fits one of our formulae. Only after trying all that do you begin the FOIL approach.

Factor 3x 2 + 14 x + 16 size 12{3x rSup { size 8{2} } +"14"x+"16"} {}

3x + __ x + __ size 12{ left (3x+"___" right ) left (x+"___" right )} {}

What multiplies to 16? 1 16 size 12{1 cdot "16"} {} , or 2 8 size 12{2 cdot 8} {} , or 4 4 size 12{4 cdot 4} {}

Which of those adds to 14 after tripling one number ? 8 + 3 2 size 12{8+3 cdot 2} {}

3x + 8 x + 2 size 12{ left (3x+8 right ) left (x+2 right )} {}

If the x 2 size 12{x rSup { size 8{2} } } {} has a coefficient, and if you can’t pull it out, the problem is trickier. In this case, we know that the factored form will look like 3x + __ x + __ size 12{ left (3x+"__" right ) left (x+"__" right )} {} so we can see that, when we multiply it back, one of those numbers—the one on the right—will be tripled, before they add up to the middle term! So you have to check the number pairs to see if any work that way.

Checking your answers

There are two different ways to check your answer after factoring: multiplying back, and trying numbers.

  1. Problem : Factor 40 x 3 250 x size 12{"40"x rSup { size 8{3} } - "250"x} {}
    • 10 x 4x 25 size 12{"10"x left (4x - "25" right )} {} First, pull out the common factor
    • 10 x 2x + 5 2x 5 size 12{"10"x left (2x+5 right ) left (2x - 5 right )} {} Difference between two squares
  2. So, does 40 x 3 250 x = 10 x 2x + 5 2x 5 size 12{"40"x rSup { size 8{3} } - "250"x="10"x left (2x+5 right ) left (2x - 5 right )} {} ? First let’s check by multiplying back.
    • 10 x 2x + 5 2x 5 size 12{"10"x left (2x+5 right ) left (2x - 5 right )} {}
    • = 20 x 2 + 50 x 2x 5 size 12{ {}= left ("20"x rSup { size 8{2} } +"50"x right ) left (2x - 5 right )} {} Distributive property
    • = 40 x 3 100 x 2 + 100 x 2 250 x size 12{ {}="40"x rSup { size 8{3} } - "100"x rSup { size 8{2} } +"100"x rSup { size 8{2} } - "250"x} {} FOIL
    • = 40 x 3 250 x  ✓ size 12{ {}="40"x rSup { size 8{3} } - "250"x} {}
  3. Check by trying a number. This should work for any number. I’ll use x = 7 and a calculator.
    • 40 x 3 250 x = ? 10 x 2x + 5 2x 5 size 12{"40"x rSup { size 8{3} } - "250"x { {}={}} cSup { size 8{?} } "10"x left (2x+5 right ) left (2x - 5 right )} {}
    • 40 7 3 250 7 = ? 10 7 2 7 + 5 2 7 5 size 12{"40" left (7 right ) rSup { size 8{3} } - "250" left (7 right ) { {}={}} cSup { size 8{?} } "10" left (7 right ) left (2 cdot 7+5 right ) left (2 cdot 7 - 5 right )} {}
    • 11970 = 11970  ✓ size 12{"11970"="11970"} {}

I stress these methods of checking answers, not just because checking answers is a generally good idea, but because they reinforce key concepts. The first method reinforces the idea that factoring is multiplication done backward . The second method reinforces the idea of algebraic generalizations.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
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emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
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Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
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Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
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hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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