Polynomials

Exercise 2.1

Question 1 (i)

The graphs of y = p(x) are given in figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:

Looking at the provided graph, the graph intersects the x-axis at 3 distinct points.

Answer: This means the polynomial p(x) has no real zeroes.

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Question 1 (ii)

The graphs of y = p(x) are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:

The graph intersects the x-axis at only one point. This means the value of the polynomial becomes zero at one point only.

Answer: 1 zero

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Question 1 (iii)

The graphs of y = p(x) are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:

The graph crosses the X-axis three times.

Therefore, the number of zeroes of p(x) is 3.

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Question 1 (iv)

The graphs of y = p(x) are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:

The graph is tangent to the X-axis at one point.

That point is a double root (a repeated root).

Even though the curve touches the X-axis only once, algebraically it counts as two zeroes because it's a root of multiplicity 2.

Answer: The number of zeroes of p(x) is 2.

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Question 1 (v)

The graphs of y = p(x) are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:

The graph crosses the X-axis four times.

Each intersection with the X-axis represents a real zero (root) of the polynomial.

Therefore, the number of zeroes of p(x) is 4.

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Question 1 (vi)

The graphs of y = p(x) are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:

The graph intersects the X-axis at three distinct points.

Each intersection represents a real root (zero) of the polynomial.

Therefore, the number of zeroes of p(x) is 3.

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Exercise 2.2

Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i)

Solution:

Step 1: Find the zeroes

We are given the polynomial: x² − 2x − 8

We will factor it.

We look for two numbers that:

• Multiply to give -8 (the constant term)

• Add to give -2 (the middle term)

The numbers are -4 and +2

So we can write: x² − 2x − 8 = (x + 2)(x − 4)

Now, set each factor equal to zero:

• x + 2 = 0 → x = -2

• x − 4 = 0 → x = 4

Answer: So the zeroes are -2 and 4

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(ii) 4s² – 4s + 1

Solution:

Step 1: Find the zeroes

Use the quadratic formula: s = (-b ± √(b² – 4ac)) / 2a

Here, a = 4, b = -4, c = 1

s = (4 ± √(16 – 16)) / 8

s = (4 ± 0) / 8

s = 4 / 8

s = 1/2

Answer: So, the zeroes are s = 1/2 and s = 1/2

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(iii) 6x² – 7x – 3

Solution:

Step 1: Find the zeroes

We’ll use the quadratic formula:
x = (-b ± √(b² – 4ac)) / 2a

Here:
a = 6, b = -7, c = -3

x = (7 ± √((-7)² – 4 × 6 × (-3))) / (2 × 6)
x = (7 ± √(49 + 72)) / 12
x = (7 ± √121) / 12
x = (7 ± 11) / 12

So,

x = (7 + 11) / 12 = 18 / 12 = 3/2
x = (7 – 11) / 12 = -4 / 12 = -1/3

Answer: Zeroes are x = 3/2 and x = -1/3

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(iv) 4u² + 8u

Solution:

Step 1: Find the zeroes

Factor the expression:

4u² + 8u = 4u(u + 2)

Now set each factor equal to zero:

1. 4u = 0 → u = 0

2. u + 2 = 0 → u = -2

Answer: Zeroes are u = 0 and u = -2

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(v) t² – 15

Solution:

Step 1: Find the zeroes

This is a difference of squares:

t² – 15 = 0
⇒ t² = 15
⇒ t = √15 or t = –√15

Answer: Zeroes are t = √15 and t = –√15

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(vi) 3x² – x – 4

Solution:

Step 1: Find the zeroes

We use the quadratic formula:
x = (–b ± √(b² – 4ac)) / 2a

Here, a = 3, b = –1, c = –4

x = (1 ± √(1² – 4 × 3 × (–4))) / (2 × 3)
x = (1 ± √(1 + 48)) / 6
x = (1 ± √49) / 6
x = (1 ± 7) / 6

Now calculate both values:

x = (1 + 7) / 6

= 8 / 6

= 4 / 3

x = (1 – 7) / 6

= –6 / 6

= –1

Answer: Zeroes are: x = 4/3 and x = –1

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Exercise. 2.2

Question 2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Solution:

Let the zeroes of the quadratic polynomial be α and β.

We are given:

• Sum of zeroes = 1/4

• Product of zeroes = −1

We use the general form of a quadratic polynomial based on its zeroes:

x² − (sum of zeroes)x + product of zeroes

Now substitute the given values:

x² − (1/4)x − 1

This is a quadratic polynomial with fractional coefficient. To remove the fraction, multiply the entire expression by 4 (the denominator of 1/4):

4 × [x² − (1/4)x − 1] = 4x² − x − 4

Answer: The required quadratic polynomial is: 4x² − x − 4

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Q2

Solution:

Let the zeroes of the quadratic polynomial be two numbers whose:

• Sum = √2

• Product = 1/3

We use the standard form:

x² − (sum of zeroes)x + product of zeroes

Substitute the given values:

x² − (√2)x + 1/3

To eliminate the fraction, multiply the entire expression by 3:

3 × [x² − √2x + 1/3] = 3x² − 3√2x + 1

Answer: The required quadratic polynomial is: 3x² − 3√2x + 1

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Q2

Solution:

Let the two zeroes be: 0 and √5

We use the form: (x − zero1) (x − zero2)

Substitute the values:

(x − 0)(x − √5) = x(x − √5)

Now multiply:

x × (x − √5) = x² − x√5

Answer: The required quadratic polynomial is : x² − x√5

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Q2

Solution:

Sum of zeroes = 1

Product of zeroes = 1

The general form of a quadratic polynomial is:

x² – (sum of zeroes) x + product of zeroes

Substitute the values:

x² – 1x + 1

Final Answer: x² – x + 1

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Q2:

Solution:

Sum of zeroes = –1/4

Product of zeroes = 1/4

Using the standard form:

x² – (sum of zeroes)x + product of zeroes

Substitute the values:

x² – (–1/4)x + 1/4

x² + (1/4)x + 1/4

To remove fractions, multiply all terms by 4:

4x² + x + 1

Final Answer: 4x² + x + 1

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Q2:

Solution:

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

Here,

Sum of zeroes = 4

Product of zeroes = 1

Substitute these values into the formula: x² - 4x + 1

Answer: The required quadratic polynomial is: x² - 4x + 1