Quick Answer: A polynomial is an expression of the form \(a_n x^n + \dots + a_0\). Its degree is the highest exponent of its variable, its zeroes are values of \(x\) where \(p(x) = 0\), and Vieta's formulas connect its coefficients to the sum and product of those zeroes.
Grade 10 Mathematics · Chapter 2
Polynomials
The algebraic sentences that encode the universe — broken down into crystal-clear steps, from definition to mastery.
✦ Every problem is solved step-by-step · Foundational → Advanced · Verification included for each section
| Polynomial Type | General Form | Sum of Zeroes (\(\alpha + \beta\)) | Product of Zeroes (\(\alpha\beta\)) |
|---|---|---|---|
| Linear | \(ax + b\) | \(-\frac{b}{a}\) | N/A |
| Quadratic | \(ax^2 + bx + c\) | \(-\frac{b}{a}\) | \(\frac{c}{a}\) |
01
Classification and Degree of Polynomials
The degree is the "rank" of a polynomial — it determines its shape, its name, and its behaviour. Master this first.
P — 01
Identify the degree of \(p(x) = 4x + 2\).
Foundational
Every polynomial wears its "rank" as the highest exponent on its variable. What rank does \(4x + 2\) hold?
Solution — Step by Step
1
Locate the variable in the expression: \(x\).
2
Identify the exponent (power) on \(x\). Since \(x = x^1\), the exponent is \(1\).
3
The constant term \(2 = 2 \cdot x^0\) has exponent \(0\), which is lower.
4
The highest exponent present is \(\mathbf{1}\), so the degree is \(1\).
Final Answer The degree of \(4x+2\) is 1. This is a Linear Polynomial.
✓ Verification
A linear polynomial must match the form \(ax + b\) with \(a \neq 0\). Here \(a = 4\) and \(b = 2\) — both real numbers with \(a \neq 0\). ✔
P — 02
Classify \(y^2 - 3y + 4\) by its degree.
Easy
Three terms, one variable, multiple powers — name this polynomial correctly.
Solution — Step by Step
1
List every term: \(y^2,\ {-3y},\ 4\).
2
Extract the power of \(y\) from each term: \(2,\ 1,\ 0\).
3
The highest power is \(\mathbf{2}\), so the degree is \(2\).
4
By definition, a polynomial of degree \(2\) is called a quadratic polynomial (derived from the Latin quadratus, meaning "square").
Final Answer \(y^2 - 3y + 4\) is a Quadratic Polynomial (degree 2).
P — 03
Find the degree of \(7u^6 - \dfrac{3}{2}u^4 + 4u^2 - 8\).
Moderate
A polynomial with four terms and fractional coefficients — don't let the fractions distract you from the exponents.
Solution — Step by Step
1
Identify all exponents of \(u\) in each term: from \(7u^6\) → exponent \(6\); from \(-\frac{3}{2}u^4\) → exponent \(4\); from \(4u^2\) → exponent \(2\); from \(-8\) → exponent \(0\).
2
Note: coefficients (even fractional ones like \(\frac{3}{2}\)) do not affect the degree. Only exponents matter.
3
Compare all exponents: \(6 > 4 > 2 > 0\). The maximum is \(\mathbf{6}\).
Final Answer The degree is 6. This polynomial has no standard name beyond "degree-6 polynomial."
P — 04
Is \(\dfrac{1}{x-1}\) a polynomial? Justify fully.
Analytical
Not everything algebraic is a polynomial. Develop the instinct to tell the difference.
Solution — Step by Step
1
Recall the definition: a polynomial in \(x\) is an expression of the form \(a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\) where all exponents are non-negative integers.
2
Rewrite the expression using exponent notation: \(\dfrac{1}{x-1} = (x-1)^{-1}\).
3
The exponent \(-1\) is negative, which violates the definition. Additionally, the variable \(x\) appears in the denominator, which is explicitly excluded.
4
Therefore, \(\frac{1}{x-1}\) does not qualify as a polynomial.
Final Answer No, \(\frac{1}{x-1}\) is not a polynomial — it contains a variable in the denominator (negative exponent).
P — 05
A polynomial has degree 3, leading coefficient \(-2\), and four terms. Write its most general form and name it.
Challenging
Build a polynomial from specifications — going from properties to expression is a key algebraic skill.
Solution — Step by Step
1
A polynomial of degree 3 is called a cubic polynomial. Its general form is: \[ax^3 + bx^2 + cx + d\] where \(a, b, c, d \in \mathbb{R}\) and \(a \neq 0\).
2
The condition "four terms" means all of \(a, b, c, d\) are present and non-zero (none can be eliminated).
3
The "leading coefficient" is the coefficient of the highest-degree term (\(x^3\)). We are told it is \(-2\), so \(a = -2\).
4
Substituting: \[-2x^3 + bx^2 + cx + d\] where \(b, c, d\) are any real numbers with \(b, c, d \neq 0\). A concrete example: \(-2x^3 + 5x^2 - 3x + 7\).
Final Answer General form: \(-2x^3 + bx^2 + cx + d\) (\(b,c,d \in \mathbb{R},\; b,c,d \neq 0\)). This is a Cubic Polynomial.
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02
Values and Zeroes of a Polynomial
Substituting a number gives a value. When that value is zero, you've struck gold — you've found a zero.
P — 01
Find the value of \(p(x) = x^2 - 3x - 4\) at \(x = 2\).
Foundational
The most fundamental operation: plug in, calculate out.
Solution — Step by Step
1
Replace every occurrence of \(x\) with \(2\): \[p(2) = (2)^2 - 3(2) - 4\]
2
Evaluate the exponent: \((2)^2 = 4\). Expression becomes \(4 - 3(2) - 4\).
3
Perform multiplication: \(3 \times 2 = 6\). Expression becomes \(4 - 6 - 4\).
4
Carry out the arithmetic left to right: \(4 - 6 = -2\); then \(-2 - 4 = -6\).
Final Answer \(p(2) = \mathbf{-6}\).
✓ Verification
Re-check: \(4 - 6 - 4 = (4-4) - 6 = 0 - 6 = -6\). Arithmetic confirmed ✔
P — 02
Show that \(x = -1\) is a zero of \(p(x) = x^2 - 3x - 4\).
Easy
To prove something is a zero, you only need to show one thing: the value must equal exactly zero.
Solution — Step by Step
1
By definition, \(x = k\) is a zero if and only if \(p(k) = 0\). We need to verify \(p(-1) = 0\).
2
Substitute \(x = -1\): \[p(-1) = (-1)^2 - 3(-1) - 4\]
3
Evaluate the square: \((-1)^2 = 1\). Evaluate the product: \(3 \times (-1) = -3\), so \(-3 \times (-1) = +3\).
4
Combine: \(1 + 3 - 4 = 4 - 4 = 0\). ✓
Final Answer Since \(p(-1) = 0\), \(x = -1\) is confirmed to be a zero of the polynomial.
P — 03
Find the zero of the linear polynomial \(p(x) = 2x + 3\).
Moderate
A linear polynomial has exactly one zero. Use the zero-condition to find it algebraically.
Solution — Step by Step
1
Set the polynomial equal to zero (the definition of a zero): \[2x + 3 = 0\]
2
Isolate the \(x\)-term by subtracting 3 from both sides: \[2x = -3\]
3
Divide both sides by the coefficient of \(x\), which is 2: \[x = -\frac{3}{2}\]
4
Notice this matches the general formula: zero of \(ax + b\) is \(-\dfrac{b}{a} = -\dfrac{3}{2}\). ✓
Final Answer The zero of \(2x + 3\) is \(x = -\dfrac{3}{2}\).
P — 04
Using the standard formula, find the zero of \(ax + b\) and express it in terms of coefficients.
Analytical
Deriving the general formula from scratch — this is how mathematicians think.
Solution — Step by Step
1
Let \(k\) be the zero of \(p(x) = ax + b\). By definition, \(p(k) = 0\), so: \[ak + b = 0\]
2
Subtract \(b\) from both sides: \[ak = -b\]
3
Divide both sides by \(a\) (valid since \(a \neq 0\) for a linear polynomial): \[k = -\frac{b}{a}\]
4
Interpret: \[k = -\frac{\text{Constant term}}{\text{Coefficient of } x}\] This elegant relationship shows that the zero is entirely determined by the coefficients.
Final Answer The zero of \(ax + b\) is \(x = -\dfrac{b}{a} = -\dfrac{\text{Constant term}}{\text{Coefficient of }x}\).
P — 05
Find the zeroes of \(p(x) = x^2 - 3\) using algebraic identities.
Challenging
No middle term — but the difference-of-squares identity unlocks this beautifully.
Solution — Step by Step
1
Using the standard Difference of Squares formula: \(a^2 - b^2 = (a-b)(a+b)\).
2
Rewrite the expression to match: \(x^2 - 3 = x^2 - (\sqrt{3})^2\).
3
Apply the identity with \(a = x,\ b = \sqrt{3}\): \[x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})\]
4
Set each factor equal to zero separately:
· \(x - \sqrt{3} = 0 \Rightarrow x = \sqrt{3}\)
· \(x + \sqrt{3} = 0 \Rightarrow x = -\sqrt{3}\)
· \(x - \sqrt{3} = 0 \Rightarrow x = \sqrt{3}\)
· \(x + \sqrt{3} = 0 \Rightarrow x = -\sqrt{3}\)
Final Answer The two zeroes of \(x^2 - 3\) are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).
03
Geometrical Meaning of Zeroes
Every zero is a story told visually — it is precisely the \(x\)-coordinate where a polynomial's graph kisses (or crosses) the \(x\)-axis.
P — 01
The graph of \(y = 2x + 3\) meets the \(x\)-axis at \(\left(-\tfrac{3}{2}, 0\right)\). State the zero.
Foundational
Reading geometry to extract algebra — the foundational link between graphs and zeroes.
Solution — Step by Step
1
Recall the key principle: the zero of a polynomial is the \(x\)-coordinate of the point where its graph intersects the \(x\)-axis.
2
The intersection point is given as \(\left(-\dfrac{3}{2},\, 0\right)\).
3
The \(x\)-coordinate of this point is \(-\dfrac{3}{2}\).
Final Answer The zero of \(2x + 3\) is \(x = -\dfrac{3}{2}\).
✓ Verification
Substitute back: \(2\!\left(-\frac{3}{2}\right) + 3 = -3 + 3 = 0\). The \(y\)-value is indeed zero at this point. ✔
P — 02
A parabola cuts the \(x\)-axis at two distinct points. How many zeroes does the quadratic have?
Easy
Translating a geometric picture directly into an algebraic count.
Solution — Step by Step
1
The graph of a quadratic \(y = ax^2 + bx + c\) is a parabola.
2
Each point where the parabola intersects the \(x\)-axis corresponds to one zero of the polynomial.
3
The problem states two distinct intersection points, call them \(A\) and \(A'\). Their two \(x\)-coordinates give two distinct zeroes.
4
This is "Case (i)" from the theory — the most common scenario for a quadratic with a positive discriminant.
Final Answer The quadratic polynomial has exactly 2 distinct zeroes.
P — 03
A parabola lies entirely above the \(x\)-axis. How many zeroes does the polynomial have?
Moderate
When there is no crossing, there is no zero. Understand why geometry dictates algebra.
Solution — Step by Step
1
A zero exists geometrically only at points where the graph meets the \(x\)-axis (i.e., where \(y = 0\)).
2
If the parabola is entirely above the \(x\)-axis, every point on it has \(y > 0\). The curve never reaches \(y = 0\).
3
Therefore, there are no intersection points with the \(x\)-axis — this is "Case (iii)" from the theory.
4
Algebraically, this corresponds to the discriminant \(b^2 - 4ac < 0\) (no real roots exist).
Final Answer The polynomial has 0 zeroes (no real zeroes).
P — 04
The graph of \(y = x^3 - 4x\) crosses the \(x\)-axis at \((-2, 0)\), \((0, 0)\), and \((2, 0)\). Identify all zeroes.
Analytical
A cubic polynomial crossing the \(x\)-axis three times — extract all zeroes from the geometry.
Solution — Step by Step
1
The graph of \(y = x^3 - 4x\) is a cubic curve. Its zeroes are the \(x\)-coordinates of all points where it touches the \(x\)-axis.
2
Three intersection points are given: \((-2, 0)\), \((0, 0)\), and \((2, 0)\).
3
Extract the \(x\)-coordinates: \(-2,\ 0,\ 2\).
4
Since a cubic polynomial can have at most 3 zeroes and we have found 3 intersection points, these are all the zeroes.
Final Answer The zeroes of \(x^3 - 4x\) are \(x = -2,\; x = 0,\; x = 2\).
P — 05
A polynomial \(p(x)\) of degree \(n\) can intersect the \(x\)-axis at how many points, at most? Prove the reasoning.
Challenging
The most general theorem in this section — understand why the bound exists.
Solution — Step by Step
1
Each intersection point of \(y = p(x)\) with the \(x\)-axis corresponds to a zero of \(p(x)\).
2
A zero \(x = k\) means \((x - k)\) is a factor of \(p(x)\). If \(p(x)\) has \(m\) distinct zeroes, then \((x - k_1)(x - k_2)\cdots(x - k_m)\) divides \(p(x)\).
3
This product \((x-k_1)\cdots(x-k_m)\) is a polynomial of degree \(m\). For it to divide \(p(x)\), we need \(m \leq n\) (a polynomial of higher degree cannot divide one of lower degree and leave a polynomial quotient).
4
Therefore, the number of intersections — and thus the number of zeroes — cannot exceed \(n\).
Final Answer A polynomial of degree \(n\) has at most \(n\) zeroes, and its graph intersects the \(x\)-axis at at most \(n\) points.
04
Relationship Between Zeroes and Coefficients
The hidden harmony of algebra: you can find the sum and product of the zeroes without ever calculating them individually.
P — 01
For \(x^2 + 7x + 10\), find the sum and product of zeroes using coefficient formulas.
Foundational
Apply the Vieta's formulas directly — no factorisation needed yet.
Solution — Step by Step
1
Identify coefficients: compare with \(ax^2 + bx + c\). Here \(a = 1,\ b = 7,\ c = 10\).
2
Using the standard Vieta's Formulas for a quadratic:
\[\alpha + \beta = -\frac{b}{a} = -\frac{7}{1} = -7\]
3
\[\alpha\beta = \frac{c}{a} = \frac{10}{1} = 10\]
Final Answer Sum of zeroes \(= -7\) · Product of zeroes \(= 10\).
✓ Verification
Factorising: \(x^2 + 7x + 10 = (x+2)(x+5)\), so zeroes are \(-2\) and \(-5\).
Sum: \(-2 + (-5) = -7\) ✔ | Product: \((-2)(-5) = 10\) ✔
P — 02
Find a quadratic polynomial whose zeroes sum to \(-3\) and multiply to \(2\).
Easy
Running the formulas in reverse — given the zero properties, reconstruct the polynomial.
Solution — Step by Step
1
Any quadratic with zeroes \(\alpha\) and \(\beta\) can be written as: \[k\bigl[x^2 - (\alpha+\beta)x + \alpha\beta\bigr],\quad k \neq 0\]
2
Substitute the given values: \(\alpha+\beta = -3\) and \(\alpha\beta = 2\):
\[k\bigl[x^2 - (-3)x + 2\bigr] = k\bigl[x^2 + 3x + 2\bigr]\]
3
The simplest polynomial is obtained by choosing \(k = 1\): \[x^2 + 3x + 2\]
4
Any scalar multiple \(k(x^2 + 3x + 2)\) also works — the family of all such polynomials shares these zero properties.
Final Answer One such polynomial is \(x^2 + 3x + 2\) (and \(k(x^2+3x+2)\) for any \(k \neq 0\)).
P — 03
Verify the zero–coefficient relationship for \(p(x) = 2x^2 - 8x + 6\).
Moderate
Find the actual zeroes first, then confirm that the formulas predict the same sums and products.
Solution — Step by Step
1
Factorise: \(2x^2 - 8x + 6 = 2(x^2 - 4x + 3) = 2(x-1)(x-3)\). So the zeroes are \(\alpha = 1\) and \(\beta = 3\).
2
Identify coefficients: \(a = 2,\ b = -8,\ c = 6\).
3
Checking the sum: Actual sum \(= 1 + 3 = 4\). Formula: \(-\dfrac{b}{a} = -\dfrac{-8}{2} = \dfrac{8}{2} = 4\). ✓
4
Checking the product: Actual product \(= 1 \times 3 = 3\). Formula: \(\dfrac{c}{a} = \dfrac{6}{2} = 3\). ✓
Final Answer Relationships verified: Sum \(= 4\), Product \(= 3\). Both formulas hold exactly.
P — 04
For the cubic \(2x^3 - 5x^2 - 14x + 8\), find the sum of products of zeroes taken two at a time.
Analytical
Cubic polynomials have three zero-coefficient relationships. This problem targets the middle one.
Solution — Step by Step
1
Compare \(2x^3 - 5x^2 - 14x + 8\) with the general form \(ax^3 + bx^2 + cx + d\):
\[a = 2,\quad b = -5,\quad c = -14,\quad d = 8\]
2
Using the standard Vieta's Formulas for Cubics, the sum of products taken two at a time is:
\[\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\]
3
Substitute: \[\alpha\beta + \beta\gamma + \gamma\alpha = \frac{-14}{2} = -7\]
4
This means: if the three zeroes are \(4,\ -2,\ \frac{1}{2}\), then \((4)(-2) + (-2)(\frac{1}{2}) + (\frac{1}{2})(4) = -8 - 1 + 2 = -7\). ✓
Final Answer Sum of products of zeroes taken two at a time \(= \dfrac{c}{a} = \mathbf{-7}\).
P — 05
Verify all three cubic relationships for \(p(x) = 3x^3 - 5x^2 - 11x - 3\) given zeroes \(3,\ -1,\ -\tfrac{1}{3}\).
Challenging
The complete verification: prove all three of Vieta's relationships for a cubic from scratch.
Solution — Step by Step
1
Identify: \(a=3,\ b=-5,\ c=-11,\ d=-3\). The zeroes are \(\alpha=3,\ \beta=-1,\ \gamma=-\dfrac{1}{3}\).
2
Relationship 1 — Sum of zeroes:
\[\alpha+\beta+\gamma = 3 + (-1) + \left(-\frac{1}{3}\right) = 2 - \frac{1}{3} = \frac{5}{3}\]
Formula: \(-\dfrac{b}{a} = -\dfrac{-5}{3} = \dfrac{5}{3}\). ✓
3
Relationship 2 — Sum of pairwise products:
\[\alpha\beta + \beta\gamma + \gamma\alpha = (3)(-1) + (-1)\!\left(-\tfrac{1}{3}\right) + \left(-\tfrac{1}{3}\right)(3) = -3 + \tfrac{1}{3} - 1 = -\tfrac{11}{3}\]
Formula: \(\dfrac{c}{a} = \dfrac{-11}{3}\). ✓
4
Relationship 3 — Product of all zeroes:
\[\alpha\beta\gamma = (3)(-1)\!\left(-\tfrac{1}{3}\right) = 1\]
Formula: \(-\dfrac{d}{a} = -\dfrac{-3}{3} = 1\). ✓
Final Answer All three Vieta relationships are verified: Sum \(=\frac{5}{3}\), Pairwise Sum \(= -\frac{11}{3}\), Product \(= 1\).
