# 7.5 General strategy for factoring polynomials

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By the end of this section, you will be able to:
• Recognize and use the appropriate method to factor a polynomial completely

Before you get started, take this readiness quiz.

1. Factor ${y}^{2}-2y-24$ .
If you missed this problem, review [link] .
2. Factor $3{t}^{2}+17t+10$ .
If you missed this problem, review [link] .
3. Factor $36{p}^{2}-60p+25$ .
If you missed this problem, review [link] .
4. Factor $5{x}^{2}-80$ .
If you missed this problem, review [link] .

## Recognize and use the appropriate method to factor a polynomial completely

You have now become acquainted with all the methods of factoring that you will need in this course. (In your next algebra course, more methods will be added to your repertoire.) The figure below summarizes all the factoring methods we have covered. [link] outlines a strategy you should use when factoring polynomials.

## Factor polynomials.

1. Is there a greatest common factor?
• Factor it out.
2. Is the polynomial a binomial, trinomial, or are there more than three terms?
• If it is a binomial:
Is it a sum?
• Of squares? Sums of squares do not factor.
• Of cubes? Use the sum of cubes pattern.
Is it a difference?
• Of squares? Factor as the product of conjugates.
• Of cubes? Use the difference of cubes pattern.
• If it is a trinomial:
Is it of the form ${x}^{2}+bx+c$ ? Undo FOIL.
Is it of the form $a{x}^{2}+bx+c$ ?
• If $a$ and $c$ are squares, check if it fits the trinomial square pattern.
• Use the trial and error or “ac” method.
• If it has more than three terms:
Use the grouping method.
3. Check.
• Is it factored completely?
• Do the factors multiply back to the original polynomial?

Remember, a polynomial is completely factored if, other than monomials, its factors are prime!

Factor completely: $4{x}^{5}+12{x}^{4}$ .

## Solution

$\begin{array}{ccccccc}\text{Is there a GCF?}\hfill & & & \text{Yes,}\phantom{\rule{0.2em}{0ex}}4{x}^{4}.\hfill & & & \hfill 4{x}^{5}+12{x}^{4}\hfill \\ & & & \text{Factor out the GCF.}\hfill & & & \hfill 4{x}^{4}\left(x+3\right)\hfill \\ \text{In the parentheses, is it a binomial, a}\hfill & & & & & & \\ \text{trinomial, or are there more than three terms?}\hfill & & & \text{Binomial.}\hfill & & & \\ \phantom{\rule{1em}{0ex}}\text{Is it a sum?}\hfill & & & & & & \text{Yes.}\hfill \\ \phantom{\rule{1em}{0ex}}\text{Of squares? Of cubes?}\hfill & & & & & & \text{No.}\hfill \\ \text{Check.}\hfill & & & & & & \\ \\ \phantom{\rule{1em}{0ex}}\text{Is the expression factored completely?}\hfill & & & & & & \text{Yes.}\hfill \\ \phantom{\rule{1em}{0ex}}\text{Multiply.}\hfill & & & & & & \\ \\ \\ \phantom{\rule{2.5em}{0ex}}4{x}^{4}\left(x+3\right)\hfill & & & & & & \\ \phantom{\rule{2.5em}{0ex}}4{x}^{4}·x+4{x}^{4}·3\hfill & & & & & & \\ \phantom{\rule{2.5em}{0ex}}4{x}^{5}+12{x}^{4}\phantom{\rule{0.2em}{0ex}}✓\hfill & & & & & & \end{array}$

Factor completely: $3{a}^{4}+18{a}^{3}$ .

$3{a}^{3}\left(a+6\right)$

Factor completely: $45{b}^{6}+27{b}^{5}$ .

$9{b}^{5}\left(5b+3\right)$

Factor completely: $12{x}^{2}-11x+2$ .

## Solution Is there a GCF? No. Is it a binomial, trinomial, or are there more than three terms? Trinomial. Are a and c perfect squares? No, a = 12, not a perfect square. Use trial and error or the “ac” method. We will use trial and error here.  Check.

$\phantom{\rule{2.5em}{0ex}}\left(3x-2\right)\left(4x-1\right)$

$\phantom{\rule{2.5em}{0ex}}12{x}^{2}-3x-8x+2$

$\phantom{\rule{2.5em}{0ex}}12{x}^{2}-11x+2\phantom{\rule{0.2em}{0ex}}✓$

Factor completely: $10{a}^{2}-17a+6$ .

$\left(5a-6\right)\left(2a-1\right)$

Factor completely: $8{x}^{2}-18x+9$ .

$\left(2x-3\right)\left(4x-3\right)$

Factor completely: ${g}^{3}+25g$ .

## Solution

$\begin{array}{ccccccc}\text{Is there a GCF?}\hfill & & & \text{Yes,}\phantom{\rule{0.2em}{0ex}}g.\hfill & & & \hfill {g}^{3}+25g\hfill \\ \text{Factor out the GCF.}\hfill & & & & & & \hfill g\left({g}^{2}+25\right)\hfill \\ \text{In the parentheses, is it a binomial, trinomial,}\hfill & & & & & & \\ \text{or are there more than three terms?}\hfill & & & \text{Binomial.}\hfill & & & \\ \phantom{\rule{1em}{0ex}}\text{Is it a sum ? Of squares?}\hfill & & & \text{Yes.}\hfill & & & \text{Sums of squares are prime.}\hfill \\ \text{Check.}\hfill & & & & & & \\ \\ \phantom{\rule{1em}{0ex}}\text{Is the expression factored completely?}\hfill & & & \text{Yes.}\hfill & & & \\ \phantom{\rule{1em}{0ex}}\text{Multiply.}\hfill & & & & & & \\ \phantom{\rule{2.5em}{0ex}}g\left({g}^{2}+25\right)\hfill & & & & & & \\ \phantom{\rule{2.5em}{0ex}}{g}^{3}+25g\phantom{\rule{0.2em}{0ex}}✓\hfill & & & & & & \end{array}$

Factor completely: ${x}^{3}+36x$ .

$x\left({x}^{2}+36\right)$

Factor completely: $27{y}^{2}+48$ .

$3\left(9{y}^{2}+16\right)$

Factor completely: $12{y}^{2}-75$ .

## Solution

$\begin{array}{ccccccc}\text{Is there a GCF?}\hfill & & & \text{Yes, 3.}\hfill & & & \hfill 12{y}^{2}-75\hfill \\ \text{Factor out the GCF.}\hfill & & & & & & \hfill 3\left(4{y}^{2}-25\right)\hfill \\ \text{In the parentheses, is it a binomial, trinomial},\hfill & & & & & & \\ \text{or are there more than three terms?}\hfill & & & \text{Binomial.}\hfill & & \\ \text{Is it a sum?}\hfill & & & \text{No.}\hfill & & & \\ \text{Is it a difference? Of squares or cubes?}\hfill & & & \text{Yes, squares.}\hfill & & & \hfill 3\left({\left(2y\right)}^{2}-{\left(5\right)}^{2}\right)\hfill \\ \text{Write as a product of conjugates.}\hfill & & & & & & \hfill 3\left(2y-5\right)\left(2y+5\right)\hfill \\ \text{Check.}\hfill & & & & & & \\ \\ \phantom{\rule{1em}{0ex}}\text{Is the expression factored completely?}\hfill & & & \text{Yes.}\hfill & & & \\ \phantom{\rule{1em}{0ex}}\text{Neither binomial is a difference of}\hfill & & & & & & \\ \phantom{\rule{1em}{0ex}}\text{squares.}\hfill & & & & & & \\ \phantom{\rule{1em}{0ex}}\text{Multiply.}\hfill & & & & & & \\ \phantom{\rule{2.5em}{0ex}}3\left(2y-5\right)\left(2y+5\right)\hfill & & & & & & \\ \phantom{\rule{2.5em}{0ex}}3\left(4{y}^{2}-25\right)\hfill & & & & & & \\ \phantom{\rule{2.5em}{0ex}}12{y}^{2}-75\phantom{\rule{0.2em}{0ex}}✓\hfill & & & & & & \end{array}$

rectangular field solutions
What is this?
Donna
the proudact of 3x^3-5×^2+3 and 2x^2+5x-4 in z7[x]/ is
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Choli
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Usman
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44.19%
Scott
40.22%
Terence
44.2%
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if you want the discounted price subtract $725 from$1299. then divide the answer by $1299. you get 0.4419... but as percent you get 44.19... but to the nearest tenth... round .19 to .2 and you get 44.2% Orlando you could also just divide$725/$1299 and then subtract it from 1. then you get the same answer. Orlando p mulripied-5 and add 30 to it Tausif Reply p mulripied-5 and add30 Tausif p mulripied-5 and addto30 Tausif Can you explain further Monica Reply p mulripied-5 and add to 30 Tausif How do you find divisible numbers without a calculator? Jacob Reply TAKE OFF THE LAST DIGIT AND MULTIPLY IT 9. SUBTRACT IT THE DIGITS YOU HAVE LEFT. IF THE ANSWER DIVIDES BY 13(OR IS ZERO), THEN YOUR ORIGINAL NUMBER WILL ALSO DIVIDE BY 13!IS DIVISIBLE BY 13 BAINAMA When she graduates college, Linda will owe$43,000 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owes for one year was $1,585. What is the amount of each loan? Ariana Reply Sean took the bus from Seattle to Boise, a distance of 506 miles. If the trip took 7 2/3 hours, what was the speed of the bus? Kirisma Reply 66miles/hour snigdha How did you work it out? Esther s=mi/hr 2/3~0.67 s=506mi/7.67hr = ~66 mi/hr Orlando hello, I have algebra phobia. Subtracting negative numbers always seem to get me confused. Alicia Reply what do you need help in? Felix subtracting a negative....is adding!! Heather look at the numbers if they have different signs, it's like subtracting....but you keep the sign of the largest number... Felix for example.... -19 + 7.... different signs...subtract.... 12 keep the sign of the "largest" number 19 is bigger than 7.... 19 has the negative sign... Therefore, -12 is your answer... Felix —12 Niazmohammad Thanks Felix.l also get confused with signs. Esther Thank you for this Shatey ty Graham think about it like you lost$19 (-19), then found $7(+7). Totally you lost just$12 (-12)
Annushka
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Meg
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Meg
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168.50=R
Heather
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46
mustee
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Joseph
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Felix
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Joyce
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3
Christopher
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