Free math worksheets

- Terms are separated by addition or subtraction at the outermost level. - Keep the subtraction sign with the term that follows it. - A coefficient is the numerical factor multiplying a variable expression.

a) The three terms are \(7x^2\), \(-3x\), and \(11\). b) The coefficient of \(x^2\) is \(7\), the coefficient of \(x\) is \(-3\), and the constant term is \(11\).

a) \(7x^2, -3x, 11\) b) Coefficient of \(x^2\): \(7\); coefficient of \(x\): \(-3\); constant: \(11\).

- First look at the expression from the outside rather than expanding it. - Factors are quantities joined by multiplication at the outermost level. - Parentheses can make a multi-part expression act as one quantity.

a) The two factors are \(6\) and \((x-4)^2\). b) The grouped expression \(x-4\) is treated as one quantity and squared.

a) \(6\) and \((x-4)^2\) b) \(x-4\)

- Match each variable with the item it counts. - A coefficient tells how much one unit contributes to the total. - A constant does not change when either variable changes.

a) The term \(8p\) is the total cost, in dollars, of the \(p\) posters. b) The coefficient \(3\) is the cost in dollars of one button pack. c) The constant \(25\) is the one-time setup charge in dollars.

a) The poster cost. b) \(\$3\) per button pack. c) The \(\$25\) one-time setup charge.

- Identify operations at the outermost level before looking inside parentheses. - A term can itself be a product of several factors. - Parentheses can signal that an entire subexpression is being used as one object.

a) The two outermost terms are \(4(2x-5)^2\) and \(9\). b) Two factors of the first term are \(4\) and \((2x-5)^2\). c) The quantity \(2x-5\) is formed first and then the entire quantity is squared. It can therefore be treated as one input to the squaring operation.

a) \(4(2x-5)^2\) and \(9\) b) \(4\) and \((2x-5)^2\) c) \(2x-5\) is one grouped subexpression that is squared as a whole.

- Start at the lowest operation in the tree and follow the outputs upward. - Use parentheses when the output of one operation becomes a single input to another. - For the final terms, look only at the outermost addition or subtraction.

a) The tree first forms \(2x\), then adds \(3\), multiplies that result by \(4\), and finally subtracts \(7\). The expression is \(4(2x+3)-7\). b) The subexpression completed before multiplying by \(4\) is \(2x+3\). c) The two outermost terms are \(4(2x+3)\) and \(-7\).

a) \(4(2x+3)-7\) b) \(2x+3\) c) \(4(2x+3)\) and \(-7\)

- Separate the expression into the product of its two main factors. - Ask what multiplying an amount by \(1.03\) does to that amount. - Treat the whole power \((1.03)^t\) as one factor rather than interpreting its pieces separately first.

a) The factor \(500\) is the initial investment amount, in dollars. b) The number \(1.03\) is the yearly growth factor: after one year, the amount is multiplied by \(1.03\), corresponding to a \(3\%\) increase. c) The factor \((1.03)^t\) is the accumulated growth factor after \(t\) years. It tells how many times the initial amount is scaled after \(t\) yearly growth steps.

a) The initial \(\$500\) investment. b) A yearly growth factor of \(1.03\), corresponding to \(3\%\) growth per year. c) The total growth factor after \(t\) years.

- Build the innermost quantity before applying the square. - Keep the phrase “the square of the quantity” grouped with parentheses. - The negative constant forms its own outer term.

The quantity \(x-4\) is squared first, giving \((x-4)^2\). Multiplying by \(3\) gives the first term \(3(x-4)^2\). Adding the constant term \(-7\) gives \(3(x-4)^2-7\). Its two outermost terms are \(3(x-4)^2\) and \(-7\).

\(3(x-4)^2-7\); its terms are \(3(x-4)^2\) and \(-7\).

- In an area formula for a rectangle, the factors correspond to side lengths. - Interpret each parenthesized expression before multiplying them. - The product has different units from either individual factor.

a) The factor \(x+4\) represents one side length of the garden, in feet. b) The factor \(x-2\) represents the other side length, in feet. c) The product of the two side lengths is the garden's area, so \((x+4)(x-2)\) represents the total area in square feet.

a) One side length, \(x+4\) feet. b) The other side length, \(x-2\) feet. c) The garden's area in square feet.

- Look first at the operations outside all parentheses. - A factor can contain addition without becoming several outer terms. - Distinguish the structure of the whole expression from the structure inside one factor.

a) The student split the expression inside the parentheses even though \(x+2\) is grouped and multiplied by \(6\). Terms of the entire expression are determined by addition or subtraction at the outermost level. b) The two terms are \(6(x+2)\) and \(-5\). c) The visible factors of the first term are \(6\) and \(x+2\).

a) The student treated parts inside a grouped factor as outer terms. b) \(6(x+2)\) and \(-5\) c) \(6\) and \(x+2\)

- Read each form according to its outermost operation. - In the factored form, multiplication is the outermost operation. - Ask how multiplying \(x+3\) by \(6\) affects each part inside the parentheses.

a) The terms are \(6x\) and \(18\). The coefficient of \(x\) is \(6\), and the constant term is \(18\). b) The visible factors are \(6\) and \(x+3\). c) The factor \(6\) shows that both terms in the expanded form are multiples of \(6\): \(6x\) is \(6\) times \(x\), and \(18\) is \(6\) times \(3\).

a) Terms: \(6x\) and \(18\); coefficient \(6\); constant \(18\). b) Factors: \(6\) and \(x+3\). c) The factor \(6\) is common to both expanded terms.

- Identify which part of the expression does not depend on \(m\). - Think about what subtracting \(3\) from the trip-meter reading counts. - Treat \(m-3\) as one quantity before interpreting the multiplication by \(2.50\).

a) The constant \(4\) is a fixed \(\$4\) charge that does not depend on the mileage beyond \(3\) miles. b) The quantity \(m-3\) is the number of miles beyond the first \(3\) miles. c) The term \(2.50(m-3)\) is the charge, in dollars, for those miles beyond the first \(3\), at \(\$2.50\) per mile.

a) The fixed \(\$4\) charge. b) Miles beyond the first \(3\) miles. c) The mileage charge for those additional miles at \(\$2.50\) per mile.

- Follow each branch from its leaves to the top. - The top operation tells how the two completed branches are related. - A subexpression can contain terms and still act as one factor in a larger expression.

a) The left branch forms \(a+b\), and the right branch forms \(a-b\). The top operation multiplies these results, so the expression is \((a+b)(a-b)\). b) The two factors are \(a+b\) and \(a-b\). c) Each sum or difference is completed on its own branch before the multiplication occurs. Therefore, each entire binomial serves as one input, or one factor, in the final product.

a) \((a+b)(a-b)\) b) \(a+b\) and \(a-b\) c) Each binomial is formed first and then used as one input to the final multiplication.

- Temporarily replace every occurrence of the repeated parenthetical expression by one symbol. - Identify the outermost addition before analyzing each term internally. - Compare how the same grouped quantity is used in the two terms. - Do not expand; the goal is to reveal structure by naming a repeated entity.

a) With \(u=x+5\), the expression has the structure \(2u+3u^2\): one term is twice \(u\), and the other is three times the square of \(u\). b) The two outermost terms are \(2(x+5)\) and \(3(x+5)^2\). c) Both terms are built from the same quantity \(x+5\). Treating that quantity as one repeated entity makes the relationship between the terms visible: one uses the first power of the entity and the other uses its square.

a) \(2u+3u^2\), where \(u=x+5\) b) \(2(x+5)\) and \(3(x+5)^2\) c) The same subexpression \(x+5\) is the common building block of both terms, appearing once to the first power and once squared.