Question

find the zeroes of the polynomial of the given below​

Answer 1

$\huge{\sf{Zeroes \: of \: the \: Polynomial}}$

To find the zeroes of the polynomial, you should make the given term equal to zero.

Basically, we learn better when we have examples, let me show you one.

x+3 is the given polynomial.

To find its zero, we should make x+3 equal to 0.

Which means,

x + 3 = 0

x = -3.

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Coming to the question,

1. P(x) = x - 5

➻ x - 5 = 0

➻ x = 5.

2. Q(x) = x + 4

➻ x + 4 = 0

➻ x = -4

3. H(x) = 6x - 1

➻ 6x - 1 = 0

➻ 6x = 1

x = $\frac{1} {6}$

4. P(x) = 4x - 5

➻ 4x - 5 = 0

➻ 4x = 5.

x = $\frac {5}{4}$

5. P(x) = ax + b

➻ ax + b = 0

➻ ax = -b

x = $\frac{-b}{a}$

6. R(x) = x² - 3x.

x² - 3x = 0.

x² - x = 3

7. L(x) = x² + 2x + 1

Here, we can split the middle term.

➻ x² + x + x + 1

➻ x ( x + 1 ) + 1 ( x + 1 )

➻ ( x + 1 ) ( x + 1 ) = ( x + 1 )²

➻ x = -1

Answer 2

1. P(x) = x - 5

x - 5 = 0

x = 5.

2. Q(x) = x + 4

x + 4 = 0

x = -4

3. H(x) = 6x - 1

6x - 1 = 0

6x = 1

x = 1 / 6

4. P(x) = 4x - 5

4x - 5 = 0

4x = 5.

x = 5 / 4

5. P(x) = ax + b

ax + b = 0

ax = -b

x = - b / a

6. R(x) = x^2 - 3x.

x^2 - 3x = 0.

x^2 - x = 3

7. L(x) = x^2 + 2x + 1

x^2 + 2x + 1 = 0

x^2 + 2x = -1

x^2 + x = -1 - 2

x^2 + x = -1

8. Q(t) = 3at - 4a^2

3at - 4a^2 = 0

At - a^2 = -3 + 4

At - a^2 = 1

9. P(x) = 2 root x