If p(x) is a polynomial and \( \alpha \) is a constant then the remainder after divison by \( (x - \alpha) \) is \( p(\alpha) \).
From the Division theorem, \[ p(x) = (x-\alpha)q(x)+ r \ \] Substituting \( x = \alpha \) gives \[ p(\alpha) = (\alpha-\alpha)q(x)+ r \ \] and \( p(\alpha) = r \).
For \(p(x)= x^2-6x+10\), find the remainder on division by (x-3).
Suppose p(x) is a polynomial and \( \alpha \) is a constant. If \( (x - \alpha) \) is a factor then \( p(\alpha) = 0 \). Conversely, if \( p(\alpha) = 0 \), then \( (x - \alpha) \) is a factor.
From the Remainder theorem, if \( (x - \alpha) \) is a factor then \( p(\alpha) = 0 \). Conversely, if \( p(\alpha) = 0 \) then \[ p(x) = (x - \alpha)q(x)+ 0 \ \] and \( (x - \alpha) \) is a factor.
For \(p(x)= x^2-6x+11\), is (x-3) a factor?
For \(p(x)= x^2-6x+9\), is (x-3) a factor?
Use the above two theorems together to discover and/or prove properties about polynomials.
For \(p(x)= x^2-ax+4\), if (x-2) is a factor, find a.
Example The polynomial \[ \begin{align} p(x) &= 2x^3 + ax^2 + bx - 3 \end{align} \ \] leaves a remainder of 3 when divided by x + 2. If x - 1 is a factor, find the values of a and b.
There are two unknowns in the expression so you need to find two equations in the unknowns a and b.
From the remainder theorem, \[ \begin{align} p(-2) &= 2(-2)^3 + a(-2)^2 + b(-2) - 3 \quad \\ 3 &= -16 + 4a - 2b - 3 \\ 4a - 2b &= 22 \qquad (1) \end{align} \ \] From the factor theorem, \[ \begin{align} p(1) &= 2(1)^3 + a(1)^2 + b(1) - 3 \quad \\ 0 &= 2 + a + b - 3 \\ a + b &= 1 \qquad (2) \end{align} \ \]
Eliminate b to find a. Adding (1) and 2 x (2): \[ \begin{align} 6a &= 24 \\ a &= 4 \end{align} \ \] from (2), \[ \begin{align} 4 + b &= 1 \\ b &= 1 - 4 \\ &= -3 \end{align} \ \] So \[ p(x) = 2x^3 + 4x^2 - 3x - 3 \ \]
Guided Examples
O E Qs
One useful observation is that if the coefficients of p(x) are all integers, then all the integer zeros of p(x) divide the constant term. Because if \( \alpha \) is an integer zero of p(x) then \[ a_n\alpha^n + \cdots + a_1\alpha + a_0 = 0 \] and \[ \begin{align} a_0 &= -(a_n\alpha^n + \cdots + a_1\alpha) \\ &= -\alpha(a_n\alpha^{n-1} + \cdots + a_1) \end{align} \ \] i.e. the integer \( \alpha \) divides the constant term.
Use the Remainder and Factor theorems, combined with polynomial division, to find the factors of polynomials. The process is:
Find two integer factors of \(p(x)= x^3-8x^2+9x+18\) by testing factors of the constant term.
Example To find all the factors of \[ p(x) = x^3 + 3x^2 -9x - 27 \ \]
Try factors of the constant term: \[ \begin{align} p(1) = (1)^3 + 3(1)^2 -9(1) - 27 &= -32 \\ p(-1) = (-1)^3 + 3(-1)^2 -9(-1) - 27 &= -16 \\ p(3) = (3)^3 + 3(3)^2 -9(3) - 27 &= 0 \\ \end{align} \ \] So \((x - 3)\) is a factor. Dividing by \((x - 3)\):
\[ \begin{matrix} \qquad\ x^2 + \;\: 6x \;\: + \; 9 \\ \qquad x-3\overline{) x^3 + 3x^2 - 9x - 27} \\ \underline{\ x^3 - 3x^2} \\ \qquad\quad\ +6x^2 - 9x \\ \qquad\qquad \ \underline{+6x^2 - 18x} \\ \qquad\qquad\qquad\qquad\ +9x - 27 \\ \qquad\qquad\qquad\qquad\ \ \underline{+9x - 27} \\ \end{matrix} \ \] And \[ x^3 + 3x^2 - 9x - 27 = (x-3)(x^2 + 6x + 9) \quad \ \]
But the second factor of p(x) is a perfect square so \[ x^3 + 3x^2 - 9x - 27 = (x-3)(x + 3)^2 \ \]