Types of polynomials — by degree and terms, with factoring strategies

Question

Classify $3x^3 - 5x^2 + 2x - 7$ by degree and number of terms. Then factorise $x^3 - 6x^2 + 11x - 6$ using the factor theorem.

Solution — Step by Step

$3x^3 - 5x^2 + 2x - 7$ has:

Degree 3 (highest power of $x$ ) → it is a cubic polynomial
4 terms → it is a polynomial with four terms (sometimes called a quadrinomial, but this term is rarely used)

Classification by degree: constant (0), linear (1), quadratic (2), cubic (3), quartic (4), quintic (5).

Classification by terms: monomial (1 term), binomial (2), trinomial (3).

For $p(x) = x^3 - 6x^2 + 11x - 6$ , we try integer factors of the constant term ( $\pm 1, \pm 2, \pm 3, \pm 6$ ):

$p(1) = 1 - 6 + 11 - 6 = 0$ ✓

So $(x - 1)$ is a factor.

Dividing $x^3 - 6x^2 + 11x - 6$ by $(x - 1)$ :

x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)

Now factorise $x^2 - 5x + 6$ : we need two numbers that multiply to 6 and add to $-5$ . Those are $-2$ and $-3$ .

x^3 - 6x^2 + 11x - 6 = \mathbf{(x-1)(x-2)(x-3)}

Why This Works

graph TD
    A["Polynomial Factoring Strategy"] --> B["Degree 2: Quadratic"]
    A --> C["Degree 3: Cubic"]
    A --> D["Degree 4+: Higher"]
    B --> E["Split middle term or use formula"]
    C --> F["Try factor theorem: find root by trial"]
    C --> G["Divide by x - root to get quadratic"]
    G --> E
    D --> H["Try grouping or repeated factor theorem"]

The factor theorem states: if $p(a) = 0$ , then $(x - a)$ is a factor of $p(x)$ . For cubics with integer coefficients, the rational root theorem tells us to try factors of the constant term divided by factors of the leading coefficient. Once we find one root, we reduce to a quadratic which we can factorise using standard methods.

For the classification: the degree determines the maximum number of roots (a cubic has at most 3 roots) and the general shape of the graph (a cubic has one “S-curve” shape with possible local max and min).

Alternative Method

For CBSE Class 10, the splitting the middle term method is the go-to for quadratic factorisation. For $ax^2 + bx + c$ , find two numbers whose product is $ac$ and sum is $b$ . Split $bx$ into these two parts, then factor by grouping.

Example: $6x^2 + 7x - 3$ . Product $= 6 \times (-3) = -18$ . Sum $= 7$ . Numbers: $9$ and $-2$ . Split: $6x^2 + 9x - 2x - 3 = 3x(2x+3) - 1(2x+3) = (3x-1)(2x+3)$ .

Common Mistake

Trying only positive integers in the factor theorem. The roots can be negative too. If $p(1), p(2), p(3)$ all give non-zero values, try $p(-1), p(-2), p(-3)$ . Also, students sometimes forget to check ALL factors of the constant term — skipping a value and then concluding “the polynomial cannot be factorised” prematurely.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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