11.3 Polynomials (Polinomlar), Quadratic Equations, Vieta's Formulas & Parabolas

Key Takeaways

  • An expression P(x) = Σ a_k x^k is a polynomial if and only if every exponent k of x is a natural number (k ∈ ℕ = {0, 1, 2, ...}); for m = deg P > n = deg Q, deg(P · Q) = m + n and deg(P^k(x^r)) = k · r · m.

  • In any polynomial expression Q(u(x)), always substitute x = 1 to find the sum of the coefficients and x = 0 to find the constant term, then evaluate the resulting input inside Q.

  • By the Remainder Theorem, dividing P(x) by ax + b leaves remainder P(-b/a), dividing by x^n - c replaces every x^n with c, and dividing by a quadratic (x - r_1)(x - r_2) leaves a linear remainder R(x) = mx + n found from P(r_1) and P(r_2).

  • For ax^2 + bx + c = 0 (a ≠ 0), Vieta's formulas give x_1 + x_2 = -b/a, x_1 · x_2 = c/a, and |x_1 - x_2| = √Δ / |a| where Δ = b^2 - 4ac; any quadratic with root sum T and root product Ç is x^2 - Tx + Ç = 0.

  • A parabola y = ax^2 + bx + c has vertex T(r, k) with r = -b/(2a) = (x_1 + x_2)/2 and k = f(r) = (4ac - b^2)/(4a), and can be written in standard form, vertex form y = a(x - r)^2 + k, or factored root form y = a(x - x_1)(x - x_2).

Last updated: October 2026

Polynomials (Polinomlar), Quadratic Equations, Vieta's Formulas & Parabolas

Polynomials, quadratic equations (ikinci dereceden denklemler), and parabolas (parabol) form a tightly connected triad on the TR-YÖS. Remainder problems turn into one-line evaluations via the Remainder Theorem, root expressions simplify without ever using the quadratic formula via Vieta's relations, and parabola geometry reduces to vertex symmetry and discriminant tangency.


1. Polynomials (Polinomlar): Exponents, Degrees & Coefficient Rules

An algebraic expression P(x)=anxn+an−1xn−1+⋯+a1x+a0P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 (ak∈Ra_k \in \mathbb{R}) is a polynomial if and only if every exponent of xx with a nonzero coefficient belongs to the natural numbers N={0,1,2,3,… }\mathbb{N} = \{0, 1, 2, 3, \dots\}.

  • Exam Trap: If P(x)=3xn−4+2x12n−5P(x) = 3x^{n-4} + 2x^{\frac{12}{n}} - 5 is a polynomial, you must enforce both n−4≥0n - 4 \ge 0 and 12n∈N\frac{12}{n} \in \mathbb{N}, which restricts nn to divisors of 1212 that are at least 44: n∈{4,6,12}n \in \{4, 6, 12\}.

Degree Rules (Derece Kuralları)

Let deg⁡[P(x)]=m\deg[P(x)] = m and deg⁡[Q(x)]=n\deg[Q(x)] = n with m>nm > n:

  • deg⁡[P(x)±Q(x)]=m\deg[P(x) \pm Q(x)] = m (if m=nm = n, deg⁡[P±Q]≤m\deg[P \pm Q] \le m because leading terms may cancel)
  • deg⁡[P(x)⋅Q(x)]=m+n\deg[P(x) \cdot Q(x)] = m + n
  • deg⁡[P(x)Q(x)]=m−n\deg\left[\frac{P(x)}{Q(x)}\right] = m - n (when Q(x)Q(x) divides P(x)P(x))
  • deg⁡[Pk(xr)]=k⋅r⋅m\deg[P^k(x^r)] = k \cdot r \cdot m and deg⁡[P(Q(x))]=m⋅n\deg[P(Q(x))] = m \cdot n

Coefficient Sums (Katsayılar Toplamı) & Constant Term (Sabit Terim)

For any polynomial expression (whether P(x)P(x), P(2x−3)P(2x - 3), or Q(x+1)Q(x + 1)):

  1. Sum of Coefficients (Katsayılar Toplamı): Substitute x=1x = 1 into the polynomial explicitly named in the question! For example, the sum of the coefficients of P(2x−3)P(2x - 3) is P(2(1)−3)=P(−1)P(2(1) - 3) = P(-1).
  2. Constant Term (Sabit Terim): Substitute x=0x = 0 into the polynomial explicitly named in the question! For example, the constant term of P(2x−3)P(2x - 3) is P(2(0)−3)=P(−3)P(2(0) - 3) = P(-3).
  3. Even- and Odd-Degree Coefficient Sums: For P(x)P(x) itself:
Even-degree coefficients=P(1)+P(−1)2,Odd-degree coefficients=P(1)−P(−1)2\text{Even-degree coefficients} = \frac{P(1) + P(-1)}{2}, \qquad \text{Odd-degree coefficients} = \frac{P(1) - P(-1)}{2}

2. Polynomial Division & the Remainder Theorem (Kalan Bulma)

In polynomial division P(x)=D(x)⋅Q(x)+R(x)P(x) = D(x) \cdot Q(x) + R(x), the remainder R(x)R(x) is either 00 or satisfies deg⁡[R(x)]<deg⁡[D(x)]\deg[R(x)] < \deg[D(x)]. You never need long division on the TR-YÖS if you set the divisor equal to zero:

  1. Linear Divisor ax+bax + b: Set ax+b=0  ⟹  x=−baax + b = 0 \implies x = -\frac{b}{a}. The remainder is the constant R=P(−ba)R = P\left(-\frac{b}{a}\right). Specifically, P(x)P(x) is divisible by x−ax - a (tam bölünür) if and only if P(a)=0P(a) = 0.
  2. Factorable Quadratic Divisor (x−r1)(x−r2)(x - r_1)(x - r_2): Since the divisor has degree 22, the remainder has at most degree 11: write R(x)=mx+nR(x) = mx + n. Substitute the two roots of the divisor:
P(r1)=mr1+nandP(r2)=mr2+nP(r_1) = m r_1 + n \quad \text{and} \quad P(r_2) = m r_2 + n

and solve the 2×22 \times 2 linear system for mm and nn. 3. Power Divisor xn−cx^n - c: Set xn−c=0  ⟹  xn=cx^n - c = 0 \implies x^n = c, and replace every power of xnx^n in P(x)P(x) with cc!


3. Quadratic Equations & Vieta's Formulas (Kök-Katsayı İlişkileri)

For the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 (a≠0a \ne 0), the discriminant is Δ=b2−4ac\Delta = b^2 - 4ac:

  • Δ>0\Delta > 0: Two distinct real roots x1,2=−b±Δ2ax_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}.
  • Δ=0\Delta = 0: One double/coincident real root (çakışık / çift katlı kök) x1=x2=−b2ax_1 = x_2 = -\frac{b}{2a}; ax2+bx+cax^2 + bx + c is a scalar multiple of a squared linear factor.
  • Δ<0\Delta < 0: No real roots (reel kök yoktur); two complex conjugate roots.

Vieta's Formulas (Kökler Toplamı, Çarpımı ve Farkı)

Never solve for x1x_1 and x2x_2 individually when asked for a symmetric combination of the roots:

T=x1+x2=−ba,C\c=x1⋅x2=ca,∣x1−x2∣=Δ∣a∣T = x_1 + x_2 = -\frac{b}{a}, \qquad Ç = x_1 \cdot x_2 = \frac{c}{a}, \qquad \lvert x_1 - x_2 \rvert = \frac{\sqrt{\Delta}}{\lvert a \rvert}
  • Essential Symmetric Identities:
    • 1x1+1x2=x1+x2x1x2=−bc\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = -\frac{b}{c}
    • x12+x22=(x1+x2)2−2x1x2=T2−2C\cx_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = T^2 - 2Ç
    • x13+x23=(x1+x2)3−3x1x2(x1+x2)=T3−3TC\cx_1^3 + x_2^3 = (x_1 + x_2)^3 - 3x_1 x_2(x_1 + x_2) = T^3 - 3TÇ
  • Constructing a Quadratic from Its Roots: Any quadratic equation whose roots have sum T=x1+x2T = x_1 + x_2 and product C\c=x1x2Ç = x_1 x_2 is written directly as:
x2−Tx+C\c=0x^2 - Tx + Ç = 0

4. Parabolas (Parabol — y=ax2+bx+cy = ax^2 + bx + c)

The graph of f(x)=ax2+bx+cf(x) = ax^2 + bx + c (a≠0a \ne 0) is a parabola opening upward (kollar yukarı, minimum at vertex) if a>0a > 0 and downward (kollar aşağı, maximum at vertex) if a<0a < 0.

Vertex (Tepe Noktası T(r,k)T(r, k)) & Axis of Symmetry

r=−b2a=x1+x22,k=f(r)=4ac−b24a=−Δ4ar = -\frac{b}{2a} = \frac{x_1 + x_2}{2}, \qquad k = f(r) = \frac{4ac - b^2}{4a} = -\frac{\Delta}{4a}

The vertical line x=rx = r is the axis of symmetry (simetri ekseni): any two points on the parabola with the same yy-value, f(p)=f(q)f(p) = f(q) (p≠qp \ne q), satisfy p+q2=r\frac{p + q}{2} = r.

Three Forms of a Parabola Equation

  1. Standard Form: y=ax2+bx+cy = ax^2 + bx + c (use when given 33 arbitrary points including the yy-intercept (0,c)(0, c)).
  2. Vertex Form (Tepe Noktası Bilinen Parabol): Given the vertex T(r,k)T(r, k) and one additional point:
y=a(x−r)2+ky = a(x - r)^2 + k
  1. Factored / Root Form (Eksenleri Kestiği Noktalar Bilinen Parabol): Given the two xx-intercepts (x1,0)(x_1, 0) and (x2,0)(x_2, 0):
y=a(x−x1)(x−x2)y = a(x - x_1)(x - x_2)

Line-Parabola Intersection

To find how the parabola y=ax2+bx+cy = ax^2 + bx + c and the line y=mx+ny = mx + n intersect, equate them to form the common equation (ortak denklem) ax2+(b−m)x+(c−n)=0ax^2 + (b - m)x + (c - n) = 0 and compute its discriminant Δcommon\Delta_{\text{common}}:

  • Δcommon>0\Delta_{\text{common}} > 0: Line intersects the parabola at 22 distinct points.
  • Δcommon=0\Delta_{\text{common}} = 0: Line is tangent (teğet) to the parabola at 11 point.
  • Δcommon<0\Delta_{\text{common}} < 0: Line and parabola do not intersect.

Summary Reference Table

TopicFormula / IdentityHigh-Speed TR-YÖS Application
Coeff Sum / ConstantPut x=1x = 1 (sum) or x=0x = 0 (constant)In P(2x−3)P(2x - 3): sum =P(−1)= P(-1), constant =P(−3)= P(-3)
Remainder TheoremP(x)÷(ax+b)  ⟹  R=P(−b/a)P(x) \div (ax + b) \implies R = P(-b/a)For P(x)÷(xn−c)P(x) \div (x^n - c), substitute xn=cx^n = c
Vieta's Formulasx1+x2=−bax_1 + x_2 = -\frac{b}{a}, x1x2=cax_1 x_2 = \frac{c}{a}, ∣x1−x2∣=Δ∣a∣\lvert x_1 - x_2 \rvert = \frac{\sqrt{\Delta}}{\lvert a \rvert}x12+x22=(x1+x2)2−2x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2
Parabola Vertex T(r,k)T(r, k)r=−b2a=x1+x22r = -\frac{b}{2a} = \frac{x_1+x_2}{2}, k=f(r)k = f(r)If f(p)=f(q)f(p) = f(q) (p≠qp \ne q), then r=p+q2r = \frac{p + q}{2}
Parabola Equationsy=a(x−r)2+ky = a(x - r)^2 + k or y=a(x−x1)(x−x2)y = a(x - x_1)(x - x_2)Line tangency: equate yparabola=yliney_{\text{parabola}} = y_{\text{line}} and set Δcommon=0\Delta_{\text{common}} = 0

Worked Examples

Worked Example 1: Coefficient Sum, Constant Term & Quadratic Divisor Remainder

Problem: For a polynomial P(x)P(x), the sum of the coefficients of P(x−1)P(x - 1) is 77, and the constant term of P(2x+3)P(2x + 3) is −5-5. Find the remainder R(x)R(x) when P(x)P(x) is divided by x2−3xx^2 - 3x.

Solution:

  1. "Sum of the coefficients of P(x−1)=7P(x - 1) = 7" means set x=1x = 1: P(1−1)=P(0)=7P(1 - 1) = P(0) = 7.
  2. "Constant term of P(2x+3)=−5P(2x + 3) = -5" means set x=0x = 0: P(2(0)+3)=P(3)=−5P(2(0) + 3) = P(3) = -5.
  3. Since the divisor x2−3x=x(x−3)x^2 - 3x = x(x - 3) has degree 22 with roots x=0x = 0 and x=3x = 3, write the remainder as R(x)=mx+nR(x) = mx + n, so P(x)=x(x−3)Q(x)+mx+nP(x) = x(x - 3)Q(x) + mx + n.
  4. Substitute x=0x = 0: P(0)=n=7P(0) = n = 7. Substitute x=3x = 3: P(3)=3m+7=−5  ⟹  3m=−12  ⟹  m=−4P(3) = 3m + 7 = -5 \implies 3m = -12 \implies m = -4. Thus R(x)=−4x+7R(x) = \mathbf{-4x + 7}.

Worked Example 2: Vieta's Symmetric Sums & Parabola Vertex/Tangency

Problem:

  1. If x1x_1 and x2x_2 are the roots of 2x2−6x−5=02x^2 - 6x - 5 = 0, find x12+x22x_1^2 + x_2^2 and 1x1+1x2\frac{1}{x_1} + \frac{1}{x_2}.
  2. A parabola y=f(x)y = f(x) has vertex T(2,−9)T(2, -9) and passes through (0,−5)(0, -5). Find the value of mm for which the line y=4x+my = 4x + m is tangent to the parabola.

Solution:

  1. By Vieta's formulas, x1+x2=−−62=3x_1 + x_2 = -\frac{-6}{2} = 3 and x1x2=−52x_1 x_2 = -\frac{5}{2}. Thus:
x12+x22=(x1+x2)2−2x1x2=32−2(−52)=9+5=14,1x1+1x2=x1+x2x1x2=3−5/2=−65x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = 3^2 - 2\left(-\frac{5}{2}\right) = 9 + 5 = \mathbf{14}, \qquad \frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = \frac{3}{-5/2} = \mathbf{-\frac{6}{5}}
  1. Using vertex form y=a(x−2)2−9y = a(x - 2)^2 - 9 and substituting (0,−5)(0, -5) gives −5=4a−9  ⟹  a=1-5 = 4a - 9 \implies a = 1, so y=(x−2)2−9=x2−4x−5y = (x - 2)^2 - 9 = x^2 - 4x - 5. Setting the parabola equal to y=4x+my = 4x + m yields the common equation x2−8x−(5+m)=0x^2 - 8x - (5 + m) = 0. For tangency, set Δcommon=0\Delta_{\text{common}} = 0:
(−8)2−4(1)(−(5+m))=64+20+4m=0  ⟹  4m=−84  ⟹  m=−21(-8)^2 - 4(1)(-(5 + m)) = 64 + 20 + 4m = 0 \implies 4m = -84 \implies m = \mathbf{-21}

Degree rules apply to nonzero polynomials and positive integer powers; the zero polynomial needs separate treatment. A quadratic divisor with distinct roots gives two independent remainder equations. For a repeated root, one evaluation is insufficient and another condition such as a derivative value is needed. Root-difference formulas above assume real roots (Δ≥0\Delta\ge0), and reciprocal-root expressions require c≠0c\ne0.

Review the Method

A parabola y = ax^2 + bx + c has vertex T(r, k) with r = -b/(2a) = (x_1 + x_2)/2 and k = f(r) = (4ac - b^2)/(4a), and can be written in standard form, vertex form y = a(x - r)^2 + k, or factored root form y = a(x - x_1)(x - x_2).

Test Your Knowledge

A polynomial P(x)P(x) satisfies P(x+2)+P(x−1)=4x+10P(x + 2) + P(x - 1) = 4x + 10 for all real xx. What is the constant term of the polynomial P(2x−1)P(2x - 1)?

A

−2-2

B

00

C

22

D

44

Test Your Knowledge

What is the remainder when the polynomial P(x)=2x15−5x10+4x5−3P(x) = 2x^{15} - 5x^{10} + 4x^5 - 3 is divided by x5−2x^5 - 2?

A

11

B

33

C

−3-3

D

55

Test Your Knowledge

If x1x_1 and x2x_2 are the roots of the quadratic equation x2−5x+m−1=0x^2 - 5x + m - 1 = 0 and satisfy 2x1−x2=12x_1 - x_2 = 1, what is the value of mm?

A

55

B

77

C

88

D

1010

Test Your Knowledge

The parabola y=−x2+6x+cy = -x^2 + 6x + c attains a maximum value of 1414 over the real numbers. What is its yy-intercept cc?

A

−5-5

B

22

C

33

D

55

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