Lesson 12

What do you notice? What do you wonder?

 

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.

\(x\)	\(x^3\)
-1

1. Finish Priya’s diagram.
2. Write \(f(x)\) as the product of \((x-1)\) and another factor.
3. Write \(f(x)\) as the product of three linear factors.
4. Make a sketch of \(y=f(x)\).

 

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.

1. \(A(x)=x^3 - 7x^2 - 16x + 112\), \((x-7)\)

   	\(x^2\)
   \(x\)	\(x^3\)	0
   -7	\(\text-7x^2\)

2. \(B(x)=2x^3 - x^2 - 27x + 36\), \(\left(x-\frac32\right)\)

   	\(2x^2\)
   \(x\)	\(2x^3\)	\(2x^2\)
   \(\text-\frac32\)	\(\text-3x^2\)

3. \(C(x)=x^3 - 3x^2 - 13x + 15\), \((x+3)\)

   \(x\)
   3

4. \(D(x)=x^4 - 13x^2 + 36\), \((x-2)\), \((x+2)\)

   (Hint: \(x^4 - 13x^2 + 36 = x^4 +0x^3 - 13x^2 +0x + 36\))

5. \(F(x)=4x^4 - 15x^3 - 48x^2 + 109x + 30\), \((x-5)\), \((x-2)\), \((x+3)\)

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.

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:

	\(x^2\)	\(\text-7x\)	+12
\(x\)	\(x^3\)	\(\text-7x^2\)	\(+12x\)
+2	\(+2x^2\)	\(\text-14x\)	+24

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 \(p(x) = (x+2)(x-3)(x-4)\).