QuanticaAlgebra IOpen in the courseChapter 1 ended on a promise. 1.4 called two expressions equivalent when they agree at every value, and said the rules for rewriting one form into another would arrive in Chapter 2. This lesson delivers the first one, and it begins as a way to multiply ugly numbers in your head.
Problem
Compute \(9\cdot 204\) in your head. The number \(204\) splits into \(200+4\), and each piece is easy to multiply by \(9\). Multiply both pieces, then add. What do you get?
Show a hint
- Multiply the easy pieces, \(9\cdot 200\) and \(9\cdot 4\), then add the two results.
Show the full solution
\(9\cdot 200=1800\) and \(9\cdot 4=36\), so the total is \(1800+36=\boxed{1836}\). Splitting a number into easy pieces beats grinding through it digit by digit.
Problem
Compute \(6\cdot 398\) the same way, but treat \(398\) as \(400-2\). Multiply each piece by \(6\), then correct for the extra. What is the product?
Show a hint
- Round \(398\) up to \(400\), multiply, then pay back the \(6\cdot 2\) you borrowed.
Show the full solution
\(6\cdot 400=2400\) and \(6\cdot 2=12\), so the product is \(2400-12=\boxed{2388}\). The correction comes off rather than on, since \(398\) sits \(2\) below \(400\).
Problem
Nothing in the rule requires the inside pieces to be plain numbers. Expand \(7(n+6)\). The result has two terms, one with \(n\) and one without. Enter the constant term of the result, the term with no \(n\).
Show a hint
- The \(7\) multiplies both pieces inside, the \(n\) and the \(6\), not just the first.
Show the full solution
Distributing to both pieces gives \(7\cdot n+7\cdot 6=7n+42\), so the term with no \(n\) is \(\boxed{42}\). Writing \(7n+6\), with the \(7\) multiplying only the first piece, is the classic mistake.
Problem
Expand \(5(3w-4)\). The first inside piece already carries a coefficient, and the second carries a minus sign. Enter the coefficient of \(w\) in the result.
Show a hint
- The minus belongs to the \(4\). Multiply \(5\) by \(3w\), then \(5\) by \(-4\), keeping that sign.
Show the full solution
\(5\cdot 3w=15w\) and \(5\cdot(-4)=-20\), so the result is \(15w-20\) and the coefficient of \(w\) is \(\boxed{15}\). The minus stays attached to the \(4\), which is what keeps the constant at \(-20\).
Problem
Expand \(g(g+11)\). The outside factor is now the variable itself, and 1.5 already settled what \(g\cdot g\) is. Enter the coefficient of the plain-\(g\) term of the result, not the \(g^2\) term.
Show a hint
- By 1.5, \(g\cdot g=g^2\). The other product is \(g\cdot 11\).
Show the full solution
Distributing gives \(g\cdot g+g\cdot 11=g^2+11g\). The plain-\(g\) term is \(11g\), so its coefficient is \(\boxed{11}\). The \(g^2\) term has a coefficient too, namely \(1\), but it belongs to the other term.
Problem
A student expands \(8(y+3)\) and writes \(8y+3\), multiplying only the first term inside. Evaluate both expressions at \(y=2\), each on its own. How far apart are their values?
Show a hint
- Substitute \(y=2\) into \(8(y+3)\) and into \(8y+3\) separately, then compare the two numbers.
Show the full solution
At \(y=2\), the original is \(8(2+3)=8\cdot 5=40\), while the student's version is \(8\cdot 2+3=19\), so the gap is \(40-19=\boxed{21}\). By 1.4, one input where two expressions disagree is enough to show they are not equivalent.
Problem
A bare minus sign directly in front of parentheses means multiplication by \(-1\), so \(-(4d-7)\) means \((-1)(4d-7)\). Rewrite it by distributing that \(-1\) over both inside terms. Enter the constant term of the result.
Show a hint
- Both inside terms get multiplied by \(-1\), so both signs flip, not just the first.
Show the full solution
Distributing hits both terms, \((-1)(4d)=-4d\) and \((-1)(-7)=+7\), so the result is \(-4d+7\) and the constant term is \(\boxed{7}\). Flipping only the first sign, giving \(-4d-7\), is the standard error.
Problem
The rule works for any number of inside terms. Compute \(12\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\) by distributing the \(12\) over all three terms. What value comes out?
Show a hint
- Each product, \(12\cdot\frac{1}{2}\), \(12\cdot\frac{1}{3}\), and \(12\cdot\frac{1}{4}\), is a whole number.
Show the full solution
\(12\cdot\frac{1}{2}+12\cdot\frac{1}{3}+12\cdot\frac{1}{4}=6+4+3=\boxed{13}\). Adding the fractions first gives \(\frac{13}{12}\) and lands in the same place, but distributing skips the common denominator entirely.
Problem
A number trick. Think of a number, add \(5\), triple the result, subtract \(15\), then divide by \(3\). Start with \(47\) and follow the four steps in order. What number comes out?
Show a hint
- Follow the four steps one at a time, keeping each intermediate result.
- When you finish, compare the number that came out with the number that went in.
Show the full solution
The steps take \(47\) to \(52\) to \(156\) to \(141\) and back to \(\boxed{47}\). Writing the start as \(n\), the first three steps build \(3(n+5)-15=3n+15-15=3n\), and dividing by \(3\) leaves \(n\), so whatever goes in comes back out.
The first rewriting tool is in hand. Ugly products split into easy pieces, and parentheses open without changing an expression's value at any input, which is exactly what 1.4 meant by equivalent. Next, 2.2 puts evaluation to work on heavier expressions, and in 2.3 this rule runs in reverse, reading \(ab+ac\) as \(a(b+c)\), a move called factoring.
Practice these ideas
Practice
Mental math first. Compute \(8\cdot 306\) in your head by splitting \(306\) into \(300+6\) and multiplying each piece by \(8\). Enter the product.
Show the solution
A common slip is to lose the middle zero and compute \(8\cdot 36=288\), which is way too small for eight copies of a number near \(300\). Keep the pieces as \(300\) and \(6\). Then \(8\cdot 300=2400\) and \(8\cdot 6=48\), so \(8\cdot 306=2400+48=\boxed{2448}\).
Practice
Compute \(4\cdot 997\) by treating \(997\) as \(1000-3\). Multiply each piece by \(4\), then correct for the overshoot. Enter the product.
Show the solution
Over subtraction the rule reads \(4(1000-3)=4\cdot 1000-4\cdot 3\). Adding the correction instead of subtracting it gives \(4012\), the wrong direction, since \(997\) sits below \(1000\). The product is \(4000-12=\boxed{3988}\).
Practice
Expand \(9(u+7)\) by multiplying each inside term by \(9\). Enter the constant term of the result, the term with no \(u\).
Show the solution
Multiplying only the first term gives \(9u+7\), the partial-distribution error, and \(9u+7\) is not equivalent to \(9(u+7)\). The \(9\) multiplies the \(7\) as well, so the expansion is \(9u+63\) and the constant term is \(\boxed{63}\).
Practice
Expand \(6(5h-9)\), keeping each inside term's sign as you distribute. Enter the coefficient of \(h\) in the result, the number multiplying it.
Show the solution
Both products keep their signs, \(6\cdot 5h=30h\) and \(6\cdot(-9)=-54\), so the expansion is \(30h-54\). Entering \(-54\) answers the wrong part, and writing \(+54\) loses the sign. The coefficient of \(h\) is \(\boxed{30}\).
Practice
Expand \(t(t+17)\). The outside factor is a variable this time. Enter the coefficient of the plain-\(t\) term of the result, not the \(t^2\) term.
Show the solution
The outside factor multiplies both terms, \(t\cdot t=t^2\) and \(t\cdot 17=17t\), so the expansion is \(t^2+17t\). It is tempting to answer with the coefficient of \(t^2\), which is \(1\), or to type the whole term \(17t\), but the question asks only for the coefficient of the plain-\(t\) term. That term is \(17t\), so the answer is \(\boxed{17}\).
Practice
Rewrite \(-(6p-19)\) by distributing the minus. A bare minus in front of parentheses means multiplication by \(-1\). Enter the constant term of the result.
Show the solution
Flipping only the first sign gives \(-6p-19\), the usual error. Multiplying each term by \(-1\) gives \(-6p\) and \((-1)(-19)=+19\), so the result is \(-6p+19\) and the constant term is \(\boxed{19}\).
Practice
A student claims that \(5(x+8)=5x+8\). Settle it the 1.4 way, evaluate both expressions at \(x=3\). How far apart are their values?
Show the solution
At \(x=3\), \(5(x+8)=5\cdot 11=55\) while \(5x+8=15+8=23\). Entering \(55\) or \(23\) alone answers a different question, the ask is the gap between them. One disagreement proves the claim false, and the gap is \(55-23=\boxed{32}\).
Practice
Three terms and two letters this time. Expand \(5(2a+9b-7)\), multiplying every term by \(5\). Enter the coefficient of \(a\).
Show the solution
The \(5\) multiplies all three terms, giving \(10a+45b-35\). The ask names the coefficient of \(a\), so \(45\) and \(-35\) answer the wrong part. It is \(5\cdot 2=\boxed{10}\).
Practice
Compute \(15\left(\frac{2}{3}+\frac{2}{5}\right)\) by distributing the \(15\) over both fractions. Done this way, no common denominator is needed.
Show the solution
Adding the fractions first gives \(\frac{16}{15}\) and also works, but distributing skips the common denominator entirely. \(15\cdot\frac{2}{3}=10\) and \(15\cdot\frac{2}{5}=6\), so the value is \(10+6=\boxed{16}\).
Practice
Compute \(7\cdot 8.03\) by splitting \(8.03\) into \(8+0.03\). Decimals distribute the same way whole numbers do. Enter the exact product.
Show the solution
The split gives \(7\cdot 8=56\) and \(7\cdot 0.03=0.21\). Misplacing the decimal point to get \(2.1\) ruins the sum, so keep \(0.21\) and the product is \(56+0.21=\boxed{56.21}\).
Practice
Expand \(3(z^2+6z+9)\). The inside has three terms, one of them squared, so the result has three terms too. Enter the constant term.
Show the solution
Every term gets the factor, so the expansion is \(3z^2+18z+27\). Stopping after two terms and leaving the \(9\) untouched is the partial-distribution error again. The constant term is \(\boxed{27}\).
Practice
Whatever number \(k\) stands for, the expression \(4(k+6)-4k\) comes out to the same value every time. Enter that value.
Show the solution
The tempting slip is to distribute the \(4\) onto \(k\) only and write \(4k+6-4k\), which would leave \(6\). The \(4\) multiplies both terms inside the parentheses, so distributing gives \(4k+24-4k\). The \(4k\) and \(-4k\) cancel completely, leaving \(24\) no matter what \(k\) is. That is the same cancellation that forced the number trick's output earlier in the lesson. The value is \(\boxed{24}\).