12.1: Notice and Wonder: A Different Use for Diagrams (5 minutes)
The purpose of this warm-up is to elicit the idea that diagrams are a useful way to stay organized when multiplying polynomials and they can also be used to divide polynomials, which is the focus of the lesson. While students may notice and wonder many things about these equations and diagrams, the relationships between the entries in the diagram and the equations are the important discussion points.
Display the equations and diagrams for all to see. Ask students to think of at least one thing they notice and at least one thing they wonder. Give students 1 minute of quiet think time, and then 1 minute to discuss the things they notice and wonder with their partner, followed by a whole-class discussion.
What do you notice? What do you wonder?
C. \((x-2)(?)=(x^3 - x^2 - 4x + 4)\)
Ask students to share the things they noticed and wondered. Record and display their responses for all to see. If possible, record the relevant reasoning on or near the image. After all responses have been recorded without commentary or editing, ask students, “Is there anything on this list that you are wondering about now?” Encourage students to respectfully disagree, ask for clarification, or point out contradicting information.
If finishing the last diagram does not come up during the conversation, ask students to discuss how they could do so starting with the entry above \(x^3\), and why it must be \(x^2\).
12.2: Factoring with Diagrams (15 minutes)
The goal of this activity is for students to understand how a diagram is useful to organize dividing polynomials. This activity continues an idea started earlier, asking: if \(x=a\) is a zero of a polynomial function, is \((x-a)\) a factor of the expression?
After students work out what quadratic times \((x-1)\) is equivalent to the original expression, they continue on to factor the quadratic using a method of their choosing, and then make a sketch of the 3rd degree polynomial, building on skills learned in previous lessons.
Ask students to close their books or devices and display the following for all to see: ”Priya wants to sketch a graph of the polynomial \(f\) defined by \(f(x)=x^3 + 5x^2 + 2x - 8\). She notices that the coefficients of the terms of \(f(x)\) sum up to zero, which means that 1 is a zero of \(f(x)\). Is she right? Be prepared to explain her reasoning.” (Priya is right that 1 is a zero of \(f(x)\) since \(f(1)=1^3+5(1)^2+2(1)-8=0\).) Give students quiet work time and then select 2–3 to share their thinking.
Priya wants to sketch a graph of the polynomial \(f\) defined by \(f(x)=x^3 + 5x^2 + 2x - 8\). She knows \(f(1)=0\), so she suspects that \((x-1)\) could be a factor of \(x^3 + 5x^2 + 2x - 8\) and writes \(\displaystyle (x^3 + 5x^2 + 2x - 8) = (x - 1)(?x^2 + ?x + ?)\) and draws a diagram.
- Finish Priya’s diagram.
- Write \(f(x)\) as the product of \((x-1)\) and another factor.
- Write \(f(x)\) as the product of three linear factors.
- Make a sketch of \(y=f(x)\).
Students who are unsure of how to complete the diagram may benefit from looking at the completed examples from the warm-up again and seeing where each term came from. If needed, help them decide in what order the parts of the diagrams were filled in.
Some students may expect to get 0 as the last step of filling in the diagram, and think that they’ve made a mistake when they get -8. Remind these students that although 1 is a zero of the function, dividing by \((x-1)\) will not necessarily produce a 0 at any step. Instead, if \((x-1)\) is a factor, the entries in the diagram will add up to the original expression. Since the constant term of the original expression is -8, the last entry in the diagram should be -8 if \((x-1)\) is a factor.
The goal of this discussion is for students to share how they logically worked out the terms of the quadratic factor of \(f(x)\) and to make sure students understand that the fact that the diagram’s final value works out to be -8 is how we know (x-1) is a factor, which wasn’t a fact when the activity started. Here are some questions for discussion:
- “How did you figure out what term must be above \(x^3\)? Below it?” (Since \(x \boldcdot x^2=x^3\), I knew \(x^2\) must be above \(x^3\), and that means \(\text-x^2\) is below it, since \(\text-1 \boldcdot x^2 = \text-x^2\).)
- “How did you work out the next column?” (The expression for \(f(x)\) has \(5x^2\), so with \(\text-x^2\) already known, the term next to \(x^3\) must be \(6x^2\), making the term above it \(6x\), since \(x \boldcdot 6x = 6x^2\), and the term below \(6x^2\) must be \(\text-6x\), since \(\text-1 \boldcdot 6x = \text-x\).)
- “If the last term in the lower right hadn’t been -8, what would that mean?” (That \((x-1)\) isn’t a factor of \(x^3 + 5x^2 + 2x - 8\).)
Design Principle(s): Optimize output (for explanation); Cultivate conversation
Supports accessibility for: Visual-spatial processing; Conceptual processing
12.3: More Factoring with Diagrams (15 minutes)
The goal of this activity is to give students additional practice using a diagram to factor when one factor (or more) is already known. The first few questions have diagrams started to give students more examples to study (in particular, that 0 is an acceptable entry). Since the focus is on working through the logic of the polynomial division to identify other factors, graphing technology is not an appropriate tool.
Arrange students in groups of 2. Tell partners to complete the diagram and then check in with one another to make sure they agree on the terms. If partners do not agree, they should work to reach agreement before moving on to the next question. Remind students that they are asked to write their result as a product of linear factors, so finishing the diagram does not finish the question.
Design Principle(s): Support sense-making
Supports accessibility for: Memory; Organization
Here are some polynomial functions with known factors. Rewrite each polynomial as a product of linear factors. Note: you may not need to use all the columns in each diagram. For some problems, you may need to make another diagram.
\(A(x)=x^3 - 7x^2 - 16x + 112\), \((x-7)\)
\(x^2\) \(x\) \(x^3\) 0 -7 \(\text-7x^2\)
\(B(x)=2x^3 - x^2 - 27x + 36\), \(\left(x-\frac32\right)\)
\(2x^2\) \(x\) \(2x^3\) \(2x^2\) \(\text-\frac32\) \(\text-3x^2\)
\(C(x)=x^3 - 3x^2 - 13x + 15\), \((x+3)\)
\(D(x)=x^4 - 13x^2 + 36\), \((x-2)\), \((x+2)\)
(Hint: \(x^4 - 13x^2 + 36 = x^4 +0x^3 - 13x^2 +0x + 36\))
\(F(x)=4x^4 - 15x^3 - 48x^2 + 109x + 30\), \((x-5)\), \((x-2)\), \((x+3)\)
Are you ready for more?
A diagram can also be used to divide polynomials even when a factor is not linear. Suppose we know \((x^2-2x+5)\) is a factor of \(x^4+x^3-5x^2+23x-20\). We could write \((x^4+x^3-5x^2+23x-20)=(x^2-2x+5)(?x^2+?x+?)\). Make a diagram and find the missing factor.
When students need to make an additional diagram, they may mistakenly start with the original polynomial instead of the result of the previous division. Remind them of how this sort of process would work with numbers: if you want to know the prime factors of 60, you can divide 60 by 5 to get 12, and then you would divide 12, not 60, by another number to continue the factorization. When finding the factors of polynomials, we likewise divide out a linear factor and then divide what remains by another linear factor until there are only linear factors.
The purpose of this discussion is for students to see and hear the ways their classmates approached the questions in the activity. Invite 1–2 students per question to share their diagram and how they worked out the other linear factors of the original expression. If time permits, highlight how the order of division does not matter for the final result. For example, whether we start with \((x-2)\) or \((x+2)\) for \(D(x)=x^4 - 13x^2 + 36\), the result is still that \(D(x)=(x-2)(x+2)(x-3)(x+3)\).
If a linear factor of a polynomial is known, a diagram is one organization strategy to identify other factors, but many other organization strategies are possible. This is even true for polynomials of degree 2. Display the following for all to see:
The polynomial \(E(x)=\text-4x^2 - 17x - 15\) has a known linear factor of \((x+3)\). Rewrite the quadratic as two linear factors.
Ask students to consider how they can factor \(\text-4x^2 - 17x - 15\) if they don’t use the diagram. After some quiet think time, invite students to share strategies they could use, recording them for all to see. For example, students might reason that the other factor must be \((\text-4x-5)\) because \(\text-4x \boldcdot x=\text-4x^2\) and \(3 \boldcdot 5=15\). If time allows, have students work through multiple strategies for factoring the quadratic and vote for the method they prefer.
12.4: Cool-down - A Product of Linear Factors (5 minutes)
Student Lesson Summary
What are some things that could be true about the polynomial function defined by \(p(x) = x^3 -5x^2 - 2x + 24\) if we know \(p(\text-2)=0\)? If we think about the graph of the polynomial, the point \((\text-2,0)\) must be on the graph as a horizontal intercept. If we think about the expression written in factored form, \((x+2)\) could be one of the factors, since \(x+2=0\) when \(x=\text-2\). How can we figure out whether \((x+2)\) actually is a factor?
Well, if we assume \((x+2)\) is a factor, there is some other polynomial \(q(x)=ax^2+bx+c\) where \(a\), \(b\), and \(c\) are real numbers and \(p(x)=(x+2)q(x)\). (Can you see why \(q(x)\) has to have a degree of 2?) In the past, we have done things like expand \((x+2)(ax^2+bx+c)\) to find \(p(x)\). Since we already know the expression for \(p(x)\), we can instead work out the values of \(a\), \(b\), and \(c\) by thinking through the calculation.
One way to organize our thinking is to use a diagram. We first fill in \((x+2)\) and the leading term of \(p(x)\), \(x^3\). From this start, we see the leading term of \(q(x)\) must be \(x^2\), meaning \(a=1\), since \(x \boldcdot x^2 = x^3\).
We then fill in the rest of the diagram using similar thinking and paying close attention to the signs of each term. For example, we put in a \(2x^2\) in the bottom left cell because that’s the product of \(2\) and \(x^2\). But that means we need to have a \(\text-7x^2\) in the middle cell of the middle row, since that’s the only other place we will get an \(x^2\) term, and we need to get \(\text-5x^2\) once all the terms are collected. Continuing in this way, we get the completed table:
Collecting all the terms in the interior of the diagram, we see that \(x^3-5x^2-2x+24=(x+2)(x^2-7x+12)\), so \(q(x) = x^2-7x+12\). Notice that the 24 in the bottom right was exactly what we needed, and it’s how we know that \((x+2)\) is a factor of \(p(x)\). In a future lesson, we will see why this happened. With a bit more factoring, we can say that \(p(x) = (x+2)(x-3)(x-4)\).