QuanticaAlgebra IOpen in the courseThe distributive property turns \(a(b+c)\) into \(ab+ac\). This lesson runs that rule backward. Reading \(ab+ac\) as \(a(b+c)\) is called factoring, and the first problems are pure arithmetic, with no letters in them at all.
Problem
Compute \(57\cdot 69+57\cdot 31\) without working out either product. Both terms contain the factor \(57\), so pull it out front and see what the leftover sum becomes.
Show a hint
- The first term is \(57\) copies of \(69\), and the second is \(57\) more copies of \(31\). Count copies before multiplying anything.
Show the full solution
Both terms share \(57\), so pull it out front. $$57\cdot 69+57\cdot 31=57(69+31)=57\cdot 100=\boxed{5700}.$$ Counting copies beats multiplying here, since the leftovers add to a round \(100\).
Problem
Compute \(58\cdot 91-58\cdot 81\) the same way. The shared factor comes out front, and this time a subtraction is left behind it. What is the value of the whole expression?
Show a hint
- The first product is \(58\) copies of \(91\) and the second is \(58\) copies of \(81\), so subtracting leaves \(58\) copies of some difference.
Show the full solution
Pull the shared \(58\) out front and the subtraction stays inside. $$58\cdot 91-58\cdot 81=58(91-81)=58\cdot 10=\boxed{580}.$$ The long way, \(5278-4698\), reaches the same \(580\) with far more work.
Problem
Letters change nothing. The expression \(4w+36\) factors as \(4(w+c)\) for one constant \(c\). Divide each term by the shared \(4\), then enter \(c\).
Show a hint
- Both terms get divided by the \(4\), not just the first one.
Show the full solution
Divide both terms by \(4\). Since \(4w\div 4=w\) and \(36\div 4=9\), the factoring is \(4(w+9)\), so \(c=\boxed{9}\). Expanding \(4(w+9)\) lands back on \(4w+36\). Dividing only the first term gives \(4(w+36)\), which expands to \(4w+144\) instead.
Problem
Dana factors \(12h+30\) as \(2(6h+15)\), and expanding it back checks out. But inside the parentheses, \(6h\) and \(15\) still share a factor, so the job is unfinished. Enter the largest number that divides both \(12h\) and \(30\).
Show a hint
- What is the largest number that divides both \(12\) and \(30\)?
Show the full solution
Since \(\gcd(12,30)=6\), the largest number dividing both terms is \(\boxed{6}\), and the complete factoring is \(6(2h+5)\). Dana's \(2(6h+15)\) expands back correctly, so passing the round trip does not mean the job is finished. Check whether the leftovers still share a factor.
Problem
A letter can be the shared factor. Since \(y^2\) means \(y\cdot y\) (1.5), both terms of \(y^2+10y\) contain \(y\), so the expression factors as \(y(y+c)\). Enter \(c\).
Show a hint
- Divide each term by \(y\) and see what each one leaves behind.
Show the full solution
Divide each term by \(y\). Since \(y^2\div y=y\) and \(10y\div y=10\), the factoring is \(y(y+10)\), and the round trip \(y(y+10)=y^2+10y\) confirms \(c=\boxed{10}\). The terms share no number bigger than \(1\), but a shared letter is a common factor too.
Problem
Number and letter combine. Enter the greatest common factor of the two terms of \(8t^2+20t\), typed like a term, for example \(6z\).
Show a hint
- Find the number part and the letter part separately, then put them together.
Show the full solution
The number part is \(\gcd(8,20)=4\), and both terms contain one \(t\), so the greatest common factor is \(\boxed{4t}\). Pulling it out gives \(4t(2t+5)\). Answering just \(4\) or just \(t\) stops halfway, since the shared factor carries a number and a letter.
Problem
A student factors \(8n+36\) as \(4(2n+8)\). Test it the 1.4 way, evaluate the original and the claimed factoring at \(n=1\), each on its own. How far apart are their values?
Show a hint
- Work out each expression at \(n=1\) separately, then subtract the results.
Show the full solution
At \(n=1\) the original gives \(8+36=44\) and the claimed factoring gives \(4(2+8)=40\), so the two are \(\boxed{4}\) apart. Expanding catches it faster, since \(4(2n+8)=8n+32\) rather than \(8n+36\). The \(4\) does divide both terms, but the leftover constant is wrong.
Problem
Three terms this time. The expression \(9a+27b-18\) factors as \(9(a+3b+c)\), and each term keeps its sign as the \(9\) comes out. Enter \(c\), sign included.
Show a hint
- What is \(-18\) divided by \(9\), sign and all?
Show the full solution
The third term is \(-18\), and \(-18\div 9=-2\), so \(9a+27b-18=9(a+3b-2)\) and \(c=\boxed{-2}\). Each leftover keeps the sign of the term it came from, so dropping the minus would break the expansion.
Problem
In \(m(m-6)+8(m-6)\), both terms contain the whole factor \(m-6\). Pull the chunk out front exactly the way the \(57\) came out in this lesson's first problem. The result is \((m+a)(m-6)\). Enter \(a\).
Show a hint
- Treat \(m-6\) as a single object, like the shared \(57\) in the hook problem.
- The first term is \(m\) copies of \(m-6\), and the second term adds some more copies of the same thing. Count the total number of copies.
Show the full solution
Treat \(m-6\) as one object. The first term is \(m\) copies of it and the second is \(8\) copies, which is \(m+8\) copies in all, so the expression equals \((m+8)(m-6)\) and \(a=\boxed{8}\). Expanding the two products into four separate terms first would erase the shared chunk, and chapter 9 factors quadratics with this same move.
Factoring reads distribution backward, the round trip checks it, completeness finishes it, and even a whole parenthesized chunk can be the shared factor. Next, 2.4 adds and subtracts entire expressions, and the full sign flip from distributing \(-1\) in 2.1 becomes the main skill.
Practice these ideas
Practice
Compute \(34\cdot 67+34\cdot 33\) without working out either product. Factor out the shared \(34\) first and see what sum is left behind it.
Show the solution
Both terms contain \(34\), so the sum is \(34(67+33)=34\cdot 100=\boxed{3400}\). Grinding out \(2278+1122\) gets there too, just slowly.
Practice
Compute \(76\cdot 53-76\cdot 43\) by factoring first. Both products are ugly, but the difference left behind the shared factor is not.
Show the solution
Pull the shared \(76\) out front, since \(76(53-43)=76\cdot 10=\boxed{760}\). Subtracting inside the parentheses is far less work than computing \(4028-3268\).
Practice
The expression \(5q+60\) factors as \(5(q+c)\) for one constant \(c\). Divide each term by the shared \(5\), then enter \(c\).
Show the solution
Divide each term by \(5\). Since \(5q\div 5=q\) and \(60\div 5=12\), the factoring is \(5(q+12)\), which expands back to \(5q+60\), so \(c=\boxed{12}\). Writing \(5(q+60)\) divides only the first term and fails the round trip.
Practice
Enter the largest number that divides both terms of \(38x+95\). Neither number is small, so factor each one before searching for what they share.
Show the solution
Since \(38=2\cdot 19\) and \(95=5\cdot 19\), the largest common factor is \(\boxed{19}\) and the complete form is \(19(2x+5)\). Guessing small divisors stalls here, since \(2\) misses \(95\) and \(5\) misses \(38\). Factor each number first instead.
Practice
Ravi factors \(44h+66\) as \(11(4h+6)\), and the expansion checks out. Even so, \(4h\) and \(6\) still share a factor, so the factoring is not complete. Enter the largest common factor of the two original terms.
Show the solution
Since \(\gcd(44,66)=22\), the largest common factor is \(\boxed{22}\), and the complete form \(22(2h+3)\) expands back to \(44h+66\). Ravi's \(11(4h+6)\) also expands correctly, but its leftovers \(4h\) and \(6\) still share \(2\), so it is not finished.
Practice
The shared factor is a letter this time. Since \(n^2\) is \(n\cdot n\), both terms of \(n^2+13n\) contain \(n\), so the expression factors as \(n(n+c)\). Enter \(c\).
Show the solution
The terms share no number, but both contain \(n\). Dividing each by \(n\) leaves \(n+13\), and the round trip \(n(n+13)=n^2+13n\) confirms \(c=\boxed{13}\).
Practice
Number and letter both matter here. Enter the greatest common factor of the two terms of \(15u^2+25u\), typed like a term, for example \(8w\).
Show the solution
The number part is \(\gcd(15,25)=5\), and both terms contain \(u\), so the greatest common factor is \(\boxed{5u}\). Pulling it out gives \(5u(3u+5)\). Entering only \(5\) or only \(u\) stops halfway.
Practice
The expression \(26x+39y-52\) factors as \(13(2x+3y+e)\). Enter \(e\), sign included.
Show the solution
Since \(-52\div 13=-4\), the factoring is \(13(2x+3y-4)\), which expands back to \(26x+39y-52\), so \(e=\boxed{-4}\). Each leftover keeps the sign of its own term, so dropping the minus changes the expression.
Practice
Evaluate \(\dfrac{7\cdot 93+7\cdot 7}{10}\) without any long multiplication. The top is a sum of two products with a shared factor, so factor it before dividing anything.
Show the solution
Factor the top first, since \(7\cdot 93+7\cdot 7=7(93+7)=7\cdot 100=700\), and \(700\div 10=\boxed{70}\). The shared \(7\) turns two ugly products into one round number.
Practice
Both terms of \(d(d+5)+3(d+5)\) contain the whole chunk \(d+5\), so it can come out front like any shared factor. Factoring it out gives \((d+a)(d+5)\). Enter \(a\).
Show the solution
Count copies of the chunk \(d+5\). The first term is \(d\) copies and the second is \(3\) more, so there are \(d+3\) copies in all, giving \((d+3)(d+5)\) and \(a=\boxed{3}\). The tempting answer \(5\) sits inside the shared chunk, not in the leftover factor.
Practice
One more full scan, this one on \(6t^2+9t\). Enter the greatest common factor of its two terms, typed like a term. Check that the leftovers share nothing before you commit.
Show the solution
Since \(\gcd(6,9)=3\) and both terms contain \(t\), the greatest common factor is \(\boxed{3t}\) and the complete form is \(3t(2t+3)\). The leftovers \(2t\) and \(3\) share only \(1\), so nothing is left to pull. Stopping at \(3\) would leave the shared \(t\) behind.
Practice
Evaluate \(41^2+41\) by factoring before you compute anything. Done right, one clean product replaces a square plus an addition.
Show the solution
Both terms contain \(41\), so \(41^2+41=41(41+1)=41\cdot 42=\boxed{1722}\). The shape is general, since \(n^2+n\) always factors as \(n(n+1)\), a product of two consecutive whole numbers.