- Factoring Techniques
- Solving Quadratic Polynomials
- Graphical Analysis
- Application-based Problems
1. Factoring Techniques
Factoring is crucial for simplifying and solving polynomial equations.
Case 1: Common Factor
Find the greatest common factor (GCF) and factor it out.
Example: Factor 6×3+9x26x^3 + 9x^2. =3×2(2x+3).= 3x^2(2x + 3).
Case 2: Grouping
Group terms to create common factors.
Example: Factor x3+3×2+x+3x^3 + 3x^2 + x + 3. =(x3+3×2)+(x+3).= (x^3 + 3x^2) + (x + 3). =x2(x+3)+1(x+3).= x^2(x + 3) + 1(x + 3). =(x+3)(x2+1).= (x + 3)(x^2 + 1).
Case 3: Splitting the Middle Term
Split the middle term in quadratic polynomials.
Example: Factor x2−5x+6x^2 – 5x + 6.
Find two numbers that multiply to 66 and add to −5-5 (i.e., −2-2 and −3-3): =x2−2x−3x+6.= x^2 – 2x – 3x + 6. =x(x−2)−3(x−2).= x(x – 2) – 3(x – 2). =(x−2)(x−3).= (x – 2)(x – 3).
2. Solving Quadratic Polynomials
To solve ax2+bx+c=0ax^2 + bx + c = 0, use:
Method 1: Factoring
Solve x2−5x+6=0x^2 – 5x + 6 = 0: x2−5x+6=(x−2)(x−3)=0.x^2 – 5x + 6 = (x – 2)(x – 3) = 0. x=2, x=3.x = 2, \, x = 3.
Method 2: Completing the Square
Solve x2−4x+3=0x^2 – 4x + 3 = 0 by completing the square:
- Rewrite x2−4x+3x^2 – 4x + 3 as: x2−4x=−3.x^2 – 4x = -3.
- Add (−42)2=4\left(\frac{-4}{2}\right)^2 = 4 to both sides: x2−4x+4=−3+4.x^2 – 4x + 4 = -3 + 4.
- Simplify: (x−2)2=1.(x – 2)^2 = 1.
- Solve: x−2=±1 ⟹ x=3, x=1.x – 2 = \pm 1 \implies x = 3, \, x = 1.
Method 3: Quadratic Formula
For ax2+bx+c=0ax^2 + bx + c = 0: x=−b±b2−4ac2a.x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}.
Example: Solve 2×2−4x−6=02x^2 – 4x – 6 = 0: a=2, b=−4, c=−6.a = 2, \, b = -4, \, c = -6. x=−(−4)±(−4)2−4(2)(−6)2(2).x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(-6)}}{2(2)}. x=4±16+484.x = \frac{4 \pm \sqrt{16 + 48}}{4}. x=4±644.x = \frac{4 \pm \sqrt{64}}{4}. x=4±84.x = \frac{4 \pm 8}{4}. x=3, x=−1.x = 3, \, x = -1.
3. Graphical Analysis
Graphs of polynomials provide visual insights into their behavior and solutions.
Quadratic Polynomials:
- Shape: Parabola.
- Key points:
- Vertex: Turning point of the parabola.
- Zeros: Points where the parabola intersects the xx-axis.
- Direction: Opens upwards (a>0a > 0) or downwards (a<0a < 0).
Example: y=x2−4x+3y = x^2 – 4x + 3.
- Factorize: (x−1)(x−3)(x – 1)(x – 3), so zeros are x=1x = 1, x=3x = 3.
- Vertex: Midpoint of zeros: x=1+32=2, y=22−4(2)+3=−1.x = \frac{1 + 3}{2} = 2, \, y = 2^2 – 4(2) + 3 = -1.
- Plot points: (1, 0), (3, 0), (2, -1).
4. Application-Based Problems
Problem 1: Cost Optimization
A farmer builds a rectangular pen with xx-meter fencing on one side. The total area is 60 m260 \, \text{m}^2. Find xx.
Area = x⋅widthx \cdot \text{width}: 60=x⋅(x−2) ⟹ x2−2x−60=0.60 = x \cdot (x – 2) \implies x^2 – 2x – 60 = 0.
Solve using the quadratic formula: x=−(−2)±(−2)2−4(1)(−60)2(1).x = \frac{-(-2) \pm \sqrt{(-2)^2 – 4(1)(-60)}}{2(1)}. x=2±4+2402.x = \frac{2 \pm \sqrt{4 + 240}}{2}. x=2±2442.x = \frac{2 \pm \sqrt{244}}{2}. x=2±2612.x = \frac{2 \pm 2\sqrt{61}}{2}. x=1±61.x = 1 \pm \sqrt{61}.
Problem 2: Physics – Motion
The height hh of a ball is modeled by h(t)=−5t2+20t+25h(t) = -5t^2 + 20t + 25. Find the time when the ball hits the ground.
Set h(t)=0h(t) = 0: −5t2+20t+25=0.-5t^2 + 20t + 25 = 0.
Divide through by −5-5: t2−4t−5=0.t^2 – 4t – 5 = 0.
Factorize: (t−5)(t+1)=0.(t – 5)(t + 1) = 0. t=5, t=−1.t = 5, \, t = -1.
Since tt cannot be negative, t=5 secondst = 5 \, \text{seconds}.
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