Lesson 15

What do you notice? What do you wonder?

Consider the polynomial function \(f(x)=x^4 - ux^3 + 24x^2 - 32x + 16\) where \(u\) is an unknown real number. If \(x-2\) is a factor, what is the value of \(u\)? Explain how you know.

1. Which of these polynomials could have \((x-2)\) as a factor?
   1. \(A(x) = 6x^2 - 7x - 5\)
   2. \(B(x) = 3 x^2 + 15 x - 42\)
   3. \(C(x) = 2x^3 + 13x^2 + 16x + 5\)
   4. \(D(x) = 3x^3 - 2x^2 - 15x + 14\)
   5. \(E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70\)
   6. \(F(x) = x^4 + 5 x^3 - 27 x^2 - 101 x - 70\)
2. Select one of the polynomials that you said doesn’t have \((x-2)\) as a factor.
   1. Explain how you know \((x-2)\) is not a factor.
   2. If you have not already done so, divide the polynomial by \((x-2)\). What is the remainder?
3. List the remainders for each of the polynomials when divided by \((x-2)\). How do these values compare to the value of the functions at \(x=2\)?

When we use long division to divide 1573 by 12, we get a remainder of 1, so \(1573 = 12(131) + 1\). When we divide by 11 instead, we get a remainder of 0, so \(1573 = 11(143)\). A remainder of 0 means that 11 is a factor of 1573. The same thing happens with polynomials. While \((x^3 + 5x^2 + 7x + 3) \div (x+2)\) results in a remainder that is not 0, if we divide \((x+1)\) into \(x^3 + 5x^2 + 7x + 3\), we do get a remainder of 0:

\(\displaystyle \require{enclose} \begin{array}{r}  x^2+4x+3 \\ x+1 \enclose{longdiv}{x^3+5x^2+7x+3} \\    \underline{\text-x^3-x^2} \phantom{+7x+333} \\  4x^2+7x \phantom{+33}\\ \underline{\text-4x^2-4x} \phantom{+33} \\ 3x+3 \\ \underline{\text-3x-3} \\ 0 \end{array}\)

So \((x+1)\) is a factor of \(x^3 + 5x^2 + 7x + 3\).

Earlier we learned that if \((x-a)\) is a factor of a polynomial \(p(x)\), then \(p(a)=0\), meaning \(a\) is a zero of the function. It turns out that the converse is also true: if \(a\) is a zero, then \((x-a)\) is a factor.

To see that this is true, let’s think about what we know if we have a polynomial \(p(x)\) with a known zero at \(x=a\). If we divide \(p(x)\) by the linear factor \((x-a)\), then \(p(x) = (x-a)q(x)+r\), where \(r\) is the remainder and \(q(x)\) is a polynomial. Because \(a\) is a zero of the function, we know that \(p(a)=0\). This means we also know that the remainder is zero:

Lastly, even if \(a\) is not a zero of \(p\), we can figure out what the remainder will be if we divide \(p(x)\) by \((x-a)\), without having to do any division. If \(p(x) = (x-a)q(x)+r\), then \(p(a)=(a-a)q(x)+r\), so \(p(a)=r\). So the remainder after division by \((x-a)\) is \(p(a)\). This is the Remainder Theorem.