Understanding Zeroes:

The zeroes of a polynomial are the x-values where p(x) = 0.

These are the points where the graph intersects the x-axis (where y = 0).

Identifying from Given Points:

Looking at the given points:

• Point (-6, 0): Here y = 0, so x = -6 is a zero
• Point (0, -30): Here y ≠ 0, so not a zero
• Point (4, -20): Here y ≠ 0, so not a zero
• Point (6, 0): Here y = 0, so x = 6 is a zero

Formula:

If α and β are the zeroes, then the quadratic polynomial is:

p(x) = k[x² - (α + β)x + αβ]

Step 1: Find Sum of Zeroes

α + β = -52 + 52 = 0

Step 2: Find Product of Zeroes

αβ = (-52)(52)

= -52

Step 3: Form the Polynomial (taking k = 2)

p(x) = 2[x² - 0·x + (-52)]

= 2[x² - 52]

= 2x² - 5

Given Information:

Sum of zeroes = α + β = -5

Product of zeroes = αβ = 6

Formula:

For a quadratic polynomial with zeroes α and β:

p(x) = k[x² - (sum of zeroes)x + (product of zeroes)]

Solution (taking k = 1):

p(x) = x² - (-5)x + 6

p(x) = x² + 5x + 6

Verification (Factorization):

x² + 5x + 6 = (x + 2)(x + 3)

Zeroes are -2 and -3

Sum = -2 + (-3) = -5 ✓

Product = (-2)(-3) = 6 ✓

Method: Factorization by Splitting the Middle Term

Given: 3x² + 11x - 4 = 0

Step 1: Find two numbers whose:

• Product = 3 × (-4) = -12
• Sum = 11

The numbers are 12 and -1

(Because: 12 × (-1) = -12 and 12 + (-1) = 11)

Step 2: Split the middle term

3x² + 11x - 4 = 3x² + 12x - x - 4

Step 3: Group and factor

= 3x(x + 4) - 1(x + 4)

= (3x - 1)(x + 4)

Step 4: Find zeroes

Setting p(x) = 0:

(3x - 1)(x + 4) = 0

Either 3x - 1 = 0 or x + 4 = 0

x = 13 or x = -4

Given Polynomial:

p(x) = 2x² + 6x - 6

Comparing with ax² + bx + c:

a = 2, b = 6, c = -6

Step 1: Use relationships between zeroes and coefficients

Sum of zeroes: α + β = -ba = -62 = -3

Product of zeroes: αβ = ca = -62 = -3

Step 2: Simplify the required expression

1α + 1β = β + ααβ = α + βαβ

Step 3: Substitute values

1α + 1β = -3-3 = 1

Given Polynomial:

x² - 3x + 2

Step 1: Find relationships for original zeroes

Sum of zeroes: α + β = 3

Product of zeroes: αβ = 2

Step 2: Find sum of new zeroes

(2α + 1) + (2β + 1) = 2α + 2β + 2

= 2(α + β) + 2

= 2(3) + 2 = 8

Step 3: Find product of new zeroes

(2α + 1)(2β + 1) = 4αβ + 2α + 2β + 1

= 4αβ + 2(α + β) + 1

= 4(2) + 2(3) + 1

= 8 + 6 + 1 = 15

Step 4: Construct the required polynomial

Using the formula:

p(x) = x² - (sum of zeroes)x + (product of zeroes)

p(x) = x² - 8x + 15

Verification:

Original zeroes: α = 1, β = 2 (from x² - 3x + 2 = (x-1)(x-2))

New zeroes: 2(1)+1 = 3, 2(2)+1 = 5

Check: x² - 8x + 15 = (x-3)(x-5) ✓

Given Information:

Sum of zeroes: α + β = 0

Product of zeroes: αβ = -9

Part 1: Construct the Polynomial

Using the formula:

p(x) = k[x² - (α + β)x + αβ]

Taking k = 1:

p(x) = x² - 0·x + (-9)

p(x) = x² - 9

Part 2: Find the Zeroes

Setting p(x) = 0:

x² - 9 = 0

This is a difference of squares:

(x - 3)(x + 3) = 0

Therefore:

x = 3 or x = -3

Verification:

Sum: 3 + (-3) = 0 ✓

Product: 3 × (-3) = -9 ✓