Substituting a number into a polynomial, and what makes it a zero

What to assume they know

• Degree as the label that separates linear, quadratic and cubic — degree, and the general forms ax + b, ax² + bx + c
• Arithmetic with negative numbers, in particular that a negative number squared is positive and cubed is negative
• Order of operations: powers before multiplication, multiplication before subtraction
• Solving a one-step linear equation such as 2k + 3 = 0

What they should be able to do

• Compute the output of a polynomial at a given input, showing the substitution before the arithmetic
• Read and write the notation p(k), and say in words what it stands for
• State the test that decides whether a number is a zero, and apply it
• Explain why the constant term is the output at input 0
• Find the zero of any linear polynomial by solving the equation the definition produces
• Express the zero of ax + b as a ratio of the constant term to the coefficient of the variable, and check the sign
• Distinguish testing a candidate from finding a zero, and say which of the two the chapter has a general method for at this point

Where it usually goes wrong

• "p(2) means p multiplied by 2." It is a single number, the output at input 2. The bracket is not multiplication here, and this is the commonest reading error at the start of the chapter.
• "A zero of a polynomial is the number 0." The zero is the input; 0 is what comes out. Two different slots, and the word points at the first.
• "p(−1) = (−1)² − 3 × (−1) − 4 = −1 − 3 − 4." Both signs are wrong. Substitute with brackets intact, then simplify — the page itself writes the brackets before evaluating.
• "To find a zero you try numbers until one works." For a linear polynomial there is a method that always terminates in one step. The trying-numbers approach is exactly what the definition removes.
• "The zero of ax + b is b/a." The sign is the point. Track it back through ak = −b rather than memorising the ratio.
• "Every polynomial has a zero you can compute this way." What has been settled here is the linear case. The quadratic case is the open question the section closes on, and it takes the rest of the chapter.
• "p(0) has to be worked out term by term." It is the constant term, every time, and seeing that is more useful than computing it.

Questions to check understanding

• Compute the value of a stated polynomial at a stated input, including negative and fractional inputs
• Decide whether a given number is a zero of a given polynomial, showing the substitution
• Find where a linear polynomial vanishes, and check the answer by substituting it back
• Given that a stated number is a zero, find an unknown coefficient — the same test read as an equation in the coefficient
• Write down the value at 0 by inspection and justify the shortcut
• Explain why a linear polynomial cannot have two different zeroes

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated polynomials.

• The chapter's running specimen (§2.1, p. 11): p(x) = x² − 3x − 4. Every substitution in this topic is on this one polynomial, which is worth keeping throughout.
  • Verified: p(2) = 4 − 6 − 4 = −6. The page prints −6 as the outcome; the intermediate 4 − 6 − 4 is the part to show.
  • Verified: p(0) = 0 − 0 − 4 = −4. Show that the first two terms vanish because both carry an x, so what survives is the constant term. This is a general fact worth stating: the output at 0 is always the constant term.
  • Verified: p(−1) = 1 + 3 − 4 = 0. Two sign traps in one line — (−1)² is positive, and subtracting 3 × (−1) adds 3.
  • Verified: p(4) = 16 − 12 − 4 = 0.
• The pair of zeroes that falls out. Since the outputs at −1 and at 4 are both 0, those two inputs are zeroes of x² − 3x − 4. Hold on to them: the same two numbers come back in §2.2 as the places where the curve of this polynomial meets the horizontal axis (Table 2.1 and Fig. 2.2, pp. 12–13).
• The linear specimen (§2.1, p. 11): p(x) = 2x + 3. The definition turns the question into the equation 2k + 3 = 0. Verified: k = −3/2, and checking, 2 × (−3/2) + 3 = −3 + 3 = 0.
• The general linear polynomial ax + b (§2.1, p. 11). The same move gives ak + b = 0, so k = −b/a. Verified as a reading of that formula: the numerator is the constant term with its sign flipped and the denominator is the coefficient of the variable, which is how the page presents it. Note: a ≠ 0 is doing real work here — it is what lets you divide.
• A worked instance to build, not printed in the chapter. Take ax + b as 4x − 10. Verified: the zero is 10/4 = 5/2, and 4 × (5/2) − 10 = 0. Useful because the sign flip is visible: b is negative, so −b/a is positive.

Figures to have open

• A one-slot machine schematic: a number goes in on the left, the polynomial sits in the box, a number comes out on the right. Standard schematic; carries sections 1 to 5.
• A substitution panel that can show a value replacing every occurrence of the variable at once. Standard schematic.
• No graph in this topic. The picture of a zero is the next topic's argument, and showing it early costs that topic its punchline.

Where this sits in the book

• NCERT Class 10 Mathematics, Chapter 2 "Polynomials", §2.1 "Introduction", p. 11 — from the first substitution into x² − 3x − 4 down to the two questions that close the section.
• Forward pointer inside the same chapter: the two zeroes found here, −1 and 4, reappear as crossings in §2.2 (Table 2.1 and Fig. 2.2, pp. 12–13), and the question the section closes on is answered in §2.3 (pp. 18–21).