EXERCISE-2

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DO THIS

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Draw the graph of (i) y = 2x + 5, (ii) y = 2x – 5, (iii) y = 2x and find the point of intersection on x-axis. Is the x-coordinates of these points also the zero of the polynomial?

y = 2x + 5 intersecting x-axis at -2.5

zero of y= 2x+5

2x+5 = 0 ⇒ x = `-5/2`=-2.5 it is same as graph.

y = 2x - 5 intersecting x-axis at 2.5

zero of y= 2x-5

2x-5 = 0 ⇒ x = `5/2`= 2.5 it is same as graph.

y = 2x intersecting x-axis at 0

zero of y= 2x

2x = 0 ⇒ x = 0 it is same as graph.

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TRY THIS

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1. Write three polynomials that have 2 zeroes each.

Quadratic polynomial has two zeroes.

If α, β are two zeroes of a polynomial then it is like p(x)=x² -(α +β)x+ αβ

If α =2, β=3 then the polynomial is x² - 5x +6

If α =-2, β=3 then the polynomial is x² - x -6

If α =2, β=-3 then the polynomial is x² + x - 6

So, the three polynomials that have 2 zeroes are x² - 5x +6, x² - x -6, x² + x - 6

2. Write one polynomial that has one zero.

A polynomial has one zero is a linear polynomial.

p(x) = 3x+6 is a polynomial has one zero

3. How will you verify if it has only one zero.

Graph of the polynomial intersects the x- axis at only one point.

or if b²-4ac=0 of p(x)= ax²+bx+c then it has only one zero.

4. Write three polynomials that have no zeroes for x that are real numbers.

If b²-4ac <0 of p(x)= ax²+bx+c then it has no real zeroes.

If a=1,b=1 and c=1 then b²-4ac = 1-4=-3 < 0 then p(x)= x²+x+1

If a=2,b=3 and c=2 then b²-4ac = 9-16=-7 < 0 then p(x)= 2x²+3x+2

If a=1,b=2 and c=2 then b²-4ac = 4-8=-4 < 0 then p(x)= x²+2x+2

5.Find the zeroes of cubic polynomials (i) -x³ (ii) x² - x³ (iii) x³ -5x²+ 6x without drawing the graph of the polynomial.

Solution:

(i) -x³=0

x=0.So,the zero of -x³ is 0.

(ii) x² - x³=0

x²(1 - x)=0

x²=0 or (1 - x)=0

So, x=0 or x=1

The zeroes of x² - x³ is 0,1.

(iii) x³ -5x²+ 6x =0

x(x²-5x+6)=0

x=0 or x²-5x+6=0

x²-5x+6 = x²-3x - 2x +6 =0

(x-3)(x-2)=0 ⇒ x-3 = 0 or x-2=0

x= 3 or x=2

∴The zeroes of x³ -5x²+ 6x is 0,3 and 2.

6.Draw the graphs of (i) y = x² - x - 6 (ii) y = 6 -x - x² and find zeroes in each case. What do you notice?

(i) y = x² - x - 6

From the above figure the graph intersecting the x-axis at -2 and 3.

Finding the zeroes of y = x² - x - 6

let y = 0

So,x² - x - 6 = 0

x² - 3x + 2x - 6 = 0

x(x-3)+2(x-3)=0

(x-3)(x+2)=0

x-3=0 or x+2=0

`therefore`x=3 or x=-2

Zeroes of y = x² - x - 6 is 3 and -2.

We observe that the intersecting points of parabola with x-axis are the zeroes of polynomial.

(ii) y = 6 - x - x²

From the above figure the graph intersecting the x-axis at 2 and -3.

Finding the zeroes of y = 6 - x - x²

let y = 0

So,6 - x - x² = 0

6 - 3x + 2x- x² = 0

3(2- x) + x(2-x)=0

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

2-x=0 or x+3=0

`therefore`x=-3 or x=2

Zeroes of y = 6 - x - x² is -3 and 2.

We observe that the intersecting points of parabola with x-axis are the zeroes of polynomial.

Example: Find the number of zeroes of the given polynomials. And also find their values. (i) p(x) = 2x + 1 (ii) q(y) = y² -1 (iii) r(z) = z³

Solution :

We will do this without plotting the graph.

(i) p(x) = 2x + 1 is a linear polynomial. It has only one zero.

To find zeroes,

Let p(x) = 0

So, 2x+1=0

∴ x = `-1/2`

The zero of the given polynomial is `-1/2`

(ii) q(y) = y² - 1 is a quadratic polynomial.

It has atmost two zeroes.

To find zeroes, let q(y) = 0

⇒ y² -1 = 0

⇒ (y + 1) (y - 1) = 0

⇒ (y + 1) = 0 or (y - 1) = 0

⇒y = -1 or y = 1

Therefore the zeroes of the polynomial are -1 and 1.

(iii) r(z) = z³ is a cubic polynomial. It has at most three zeroes.

Let r(z) = 0

⇒ z³ = 0

⇒ z = 0

So, the zero of the polynomial is 0.

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Exercise:

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

The number of zeroes is 0 as the graph not intersects the x-axis.

The number of zeroes is 1 as the graph intersects the x-axis at one point only.

The number of zeroes is 3 as the graph intersects the x-axis at three points.

The number of zeroes is 2 as the graph intersects the x-axis at two points.

The number of zeroes is 4 as the graph intersects the x-axis at four points.

The number of zeroes is 3 as the graph intersects the x-axis at three points.

2. Find the zeroes of the given polynomials.

(i) p(x) = 3x

Let p(x) = 0

So, 3x=0

∴ x = 0

(ii) p(x) = x² + 5x + 6

Let p(x) = 0

So, x² + 5x + 6=0

x² + 3x + 2x + 6=0

x(x+2)+2(x+3)=0

(x+2) (x+3)=0

(x+2) =0 or (x+3)=0

∴ x = -2 or x=-3.

(iii) p(x) = (x+2) (x+3)

Let p(x) = 0

So, (x+2) (x+3)=0

(x+2) =0 or (x+3)=0

∴ x = -2 or x=-3.

(iv) p(x) = x⁴ -16

Let p(x) = 0

So, x⁴ -16=0

(x²)² - 4 ²=0

(x² -4)(x² +4)=0 (`because` x²-y²=(x+y)(x-y))

(x-2)(x+2)(x² +4)=0

(x-2)=0 or (x+2)=0 or (x² +4)=0

x=2 or x=-2 or x=`pm sqrt(-4)`

∴ The zeroes of P(x) is 2 ,-2 and`pm sqrt(-4)`

3. Draw the graphs of the given polynomial and find the zeroes. Justify the answers.

(i) p(x) = x² - x -12

The graph intersects the x-axis at (-3,0),(4,0).

So, the zero of the polynomial is -3,4.

Justification:

Given p(x) = x² - x -12

let p(x)=0

So, x² - x -12=0

x² - 4x + 3x +9=0

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

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

x-4=0 or x+3=0

`therefore`x=4 or x=-3

So, the zero of p(x) are 4 and -3

(ii) p(x) = x² - 6x + 9

The graph intersects the x-axis at (3,0).

So, the zero of the polynomial is 3.

Justification:

Given p(x) = x² - 6x +9

let p(x)=0

So, x² - 6x +9=0

x² - 3x - 3x +9=0

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

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

x-3=0 or x-3=0

`therefore` x=3

So, the zero of p(x) is 3

(iii) p(x) = x² - 4x + 5

The graph does not intersects the x-axis at any point.

So, the polynomial has no zeroes.

Justification:

For the given p(x), not possible to split into factors.

(iv) p(x) = x² +3x- 4

The graph intersects the x-axis at (-4,0) and (1,0).

So, the zeroes of the polynomial are -4 and 1.

Justification:

Given p(x) = x² + 3x - 4

let p(x)=0

So, x² + 3x - 4=0

x² + 4x - x - 4=0

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

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

x+4=0 or x-1=0

`therefore` x=-4 or x=1

So, the zeroes of p(x) are -4 and 1

(v) p(x) = x² - 1

The graph intersects the x-axis at (-1,0) and (1,0).

So, the zeroes of the polynomial are -1 and 1.

Justification:

Given p(x) = x² - 1

let p(x)=0

So, x² - 1=0

(x+1)(x-1)=0 [`because`a²-b² = (a+b)(a-b)]

x+1=0 or x-1=0

`therefore` x=-1 or x=1

So, the zeroes of p(x) are -1 and 1

4. Why are `1/4` and -1 zeroes of the polynomials p(x) = 4x² + 3x - 1?

Solution:

p(x) = 4x² + 3x - 1

p(`1/4`) = 4`times (1/4)^2`+ 3`times 1/4` - 1

= `1/4 + 3/4 -1`=1 -1 =0

p(-1) = 4 `times (-1)^2` + 3 `times(-1)` - 1

= 4 `times`1 - 3 -1= 4-4=0

p(`1/4`)=0 and p(-1) =0 so,`1/4` and -1 are zeroes of the polynomials p(x) = 4x² + 3x - 1

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