Remainder and Factor Theorems — Part 1: The remainder theorem and the factor theorem: visual lesson, ICSE Class 10 Mathematics

Lesson

Remainder and Factor Theorems, Part 1

Dividing a polynomial by $(x - a)$ leaves a remainder you can find without dividing: it is the value of the polynomial at $x = a$ .

In this lesson you will

Divide a polynomial by a linear expression and write dividend = divisor × quotient + remainder
Find a remainder without long division, using the remainder theorem
Find unknown coefficients when remainders are given
Use the factor theorem to test whether a linear expression is a factor

$y = x^3 - 3x^2 - 6x + 8$

The remainder on division by $(x - 3)$ is the height of the graph at $x = 3$: $f(3) = -10$.

Concept

What you will be able to do

Divide a polynomial by a linear expression
Find a remainder quickly with the remainder theorem
Find unknown coefficients from given remainders
Test factors with the factor theorem

Concept

Dividing by a linear expression

Long division of a polynomial by $(x - a)$ gives a quotient $q(x)$ and a remainder $r$ . Because the divisor has degree $1$ , the remainder is a constant.

f(x) = (x - a)\,q(x) + r

Quotient $2x^2 + x + 6$, remainder $7$: $2x^3 - 3x^2 + 4x - 5 = (x - 2)(2x^2 + x + 6) + 7$.
Step	Divide the leading terms	Multiply back	Subtract to get
1	$\dfrac{2x^3}{x} = 2x^2$	$2x^2(x - 2) = 2x^3 - 4x^2$	$x^2 + 4x - 5$
2	$\dfrac{x^2}{x} = x$	$x(x - 2) = x^2 - 2x$	$6x - 5$
3	$\dfrac{6x}{x} = 6$	$6(x - 2) = 6x - 12$	$7$

Cookies