Polynomials

The number of zeroes for any graph is the number of values of x for which y is equal to zero. And y is equal to zero at the point where a graph cuts x axis.
So, to find the number of zeroes of a polynomial, watch the number of times it cuts the x axis.
(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

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

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

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

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

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

Zeroes of the polynomial are the values of the variable of the polynomial when the polynomial is put equal to zero.
Let p(x) be a polynomial with any number of terms any number of degree.
Now, zeroes of the polynomial will be the values of x at which p(x) = 0.
If p(x) = ax² + bx + c is a quadratic polynomial (highest power is equal to 2) and its roots are α and β, then
Sum of the roots = α + β = - b/a
Product of roots = αβ = c/a

(i) p(x) = x² - 2x - 8
So, the zeroes will be the values of x at which p(x) = 0
Therefore,

⇒ x² - 4x + 2x - 8 = 0

(We will factorize 2 such that the product of the factors is equal to 8 and difference is equal to 2)
⇒ x(x - 4) + 2(x - 4) = 0
= (x - 4)(x + 2)

The value of x² - 2x - 8 is zero when x − 4 = 0 or x + 2 = 0,

i.e, x = 4 or x = −2

Therefore, The zeroes of x² - 2x - 8 are 4 and −2.

Sum of zeroes = 4 + (-2) = 2

Hence, it is verified that,

Product of zeroes =

Hence, it is verified that,

(ii) 4s² - 4s + 1
= (2s)² - 2(2s)1 + 1²
As, we know (a - b)² = a² - 2ab + b², the above equation can be written as
= (2s - 1)²

The value of 4s² − 4s + 1 is zero when 2s − 1 = 0, when, s = 1/2 , 1/2.

Therefore, the zeroes of 4s² − 4s + 1 are and .

Sum of zeroes =

Product of zeroes =

Hence Verified.

(iii)6x² - 3 - 7x
= 6x² - 7x - 3
(We will factorize 7 such that the product of the factors is equal to 18 and the difference is equal to - 7)
= 6x² + 2x - 9x - 3
= 2x(3x + 1) - 3(3x + 1)
= (3x + 1)(2x - 3)

The value of 6x² − 3 − 7x is zero when 3x + 1 = 0 or 2x − 3 = 0,

i.e.

Therefore, the zeroes of 6x² − 3 − 7x are .

Sum of zeroes =

Product of zeroes =

Hence, verified.

(iv) 4u² + 8u
= 4u² + 8u + 0

= 4u (u + 2)

The value of 4u² + 8u is zero when 4u = 0 or u + 2 = 0,

i.e., u = 0 or u = −2

Therefore, the zeroes of 4u² + 8u are 0 and −2.

Sum of zeroes =

Product of zeroes =

(v) t² – 15

= t² – (√15)²

= (t - √15)(t + √15) [As, x² - y² = (x - y)(x + y)]

The value of t² − 15 is zero when (t - √15) = 0 or (t + √15) = 0,

i.e., when t = √15 or t = -√15

Therefore, the zeroes of t² − 15 are √15 and -√15.

Sum of zeroes =

Product of zeroes =

Hence verified.

(vi) 3x² – x – 4
(We will factorize 1 in such a way that the product of factors is equal to 12 and the difference is equal to 1)
= 3x² - 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (3x – 4 )(x + 1)

The value of 3x² − x − 4 is zero when 3x − 4 = 0 or x + 1 = 0,

when

Therefore, the zeroes of 3x² − x − 4 are and -1

Sum of zeroes =

Product of zeroes =

Hence, verified.

(i) , -1

Let the polynomial be , then

Let a = 4, then b = -1, c = -4

Therefore, the quadratic polynomial is 4x² − x − 4.

(ii)

Let the polynomial be , then

If a = 3, then b = , and c = 1

Therefore, the quadratic polynomial is

(iii) 0,

Let the polynomial be , then

If a = 1, then b = 0, c =

Therefore, the quadratic polynomial is .

(iv) 1, 1

Let the polynomial be , then

If a = 1, then b = -1, c = 1

Therefore, the quadratic polynomial is .

(v)

Let the polynomial be , then

If a = 4, then b = 1, c = 1

Therefore, the quadratic polynomial is .

(vi) 4, 1

Let the polynomial be , then

If a = 1, then b = -4, c = 1

Therefore, the quadratic polynomial is .

(i)By long division method we have,

Quotient = x − 3

Remainder = 7x − 9

(ii) By long division method we have,

Quotient = x² + x − 3

Remainder = 8

(iii) By long division method we have,

Quotient = −x² − 2

Remainder = −5x +10

(i) t²-3 = t²+0t-3

Since the remainder is 0,

Hence, t² – 3 is a factor of 2t⁴+3t³-2t²-9t-12.

_(ii)

Since the remainder is 0,

Hence, x²+3x+1 is a factor of 3x⁴+5x³-7x²+2x+2.

_(iii)

Since the remainder ≠0,

Hence, x³-3x+1 is not a factor of x⁵-4x³+ x²+3x+1.

p(x) = 3x⁴ + 6x³ - 2x² - 10x - 5

Since the two zeroes are .

is a factor of 3x⁴ + 6x³ - 2x² - 10x - 5.

Therefore, we divide the given polynomial by .

We know,
Dividend = (Divisor × quotient) + remainder

3 x⁴ + 6 x³ - 2x² - 10 x - 5 =

3 x⁴ + 6 x³ - 2 x² - 10 x - 5 =
As (a+b)² = a² + b² + 2ab

3 x⁴ + 6 x³ - 2 x² - 10 x - 5 = 3 () (x + 1)²

Therefore, its zero is given by x + 1 = 0.
⇒ x = −1,-1

Hence, the zeroes of the given polynomial are and - 1 , -1.

Given,
Polynomial, p(x) = x³ - 3x² + x + 2 (dividend)

Quotient = (x − 2)

Remainder = (− 2x + 4)

⇒ x³ - 3x² + x + 2 + 2x - 4 = g(x)(x - 2)

⇒ x³ - 3x² + 3x - 2 = g(x)(x - 2)

g(x) is the quotient when we divide (x³ - 3x² + 3x - 2) by (x - 2)

Degree of a polynomial is the highest power of the variable in the polynomial. For example if f(x) = x³ - 2x² + 1, then the degree of this polynomial will be 3.

(i) By division Algorithm : p(x) = g(x) x q(x) + r(x)
It means when P(x) is divided by g(x) then quotient is q(x) and remainder is r(x)
We need to start with p(x) = q(x)
This means that the degree of polynomial p(x) and quotient q(x) is same. This can only happen if the degree of g(x) = 0 i.e p(x) is divided by a constant
Let p(x) = x² + 1 and g(x) = 2
The,

Clearly, Degree of p(x) = Degree of q(x)
2. Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

=

Thus, the division algorithm is satisfied.

(ii) Let us assume the division of x³+ x by x²,

Here,

p(x) = x³+ x

g(x) = x²

q(x) = x and r(x) = x

Clearly, the degree of q(x) and r(x) is the same i.e.,
Checking for division algorithm,

p(x) = g(x) × q(x) + r(x) x³+ x

= (x² ) × x + x x³+ x = x³+ x

Thus, the division algorithm is satisfied.

(iii) Degree of the remainder will be 0 when the remainder comes to a constant.

Let us assume the division of x³+ 1by x².

Here,

p(x) = x³+ 1 g(x) = x²

q(x) = x and r(x) = 1

Clearly, the degree of r(x) is 0. Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)x³+ 1

= (x² ) × x + 1 x³+ 1 = x³+1

Thus, the division algorithm is satisfied.

(i) P(x) =

Now for zeroes, putting the given values in x.

P(1/2) = 2(1/2)³ + (1/2)² - 5(1/2) + 2
= (1/4) + (1/4) - (5/2) + 2
= (1 + 1 - 10 + 8)/2
= 0/2 = 0

P(1) =

P(-2) =

Thus, 1/2, 1 and -2 are zeroes of given polynomial.

Comparing given polynomial with ax³ + bx² + cx + d and Taking zeroes as α, β, and γ, we have

Now, We know the relation between zeroes and the coefficient of a standard cubic polynomial as

Substituting value, we have

Since LHS = RHS, Relation verified.

Thus, all three relationships between zeroes and the coefficient is verified.

(ii) p(x) = x³ – 4x² + 5x – 2

Now for zeroes , put the given value in x.

P(2) = =

P(1) =

P(1) =

Thus, 2, 1 , 1 are the zeroes of the given polynomial.

Now,

Comparing the given polynomial with ax³ + bx² + cx + d, we get

4 = 4

5 = 5

αβγ =

2 × 1 × 1 = 2

2 = 2

Thus, all three relationships between zeroes and the coefficient is verified.

For a cubic polynomial equation, ax³ + bx² + cx + d, and zeroes α, β and γ

we know that

Let the polynomial be ax³ + bx² + cx + d, and zeroes α, β and γ.
A cubic polynomial with respect to its zeroes is given by,
x³ - (sum of zeroes) x² + (Sum of the product of roots taken two at a time) x - Product of Roots = 0
x³ - (2) x² + (- 7) x - (- 14) = 0
x³ - (2) x² + (- 7) x + 14 = 0

Hence, the polynomial is x³ - 2x² - 7x + 14.

Given,

p(x) =

zeroes are = a – b , a + b , a

C

=

Sum of zeroes =

a =

The zeroes are = (1 - b), 1 and (1 + b)

Product of zeroes = (1 - b)(1 + b)

(1 - b)(1 + b) = -q/m

So,

We get,

Given:

2+√3 and 2-√3 are zeroes of given equation,

Therefore,

(x - 2 + √3)(x - 2 - √3) should be a factor of given equation.

Also, (x - 2 + √3)(x - 2 - √3) = x² - 2x - √3x -2x + 4 + 2√3 + √3x - 2√3 - 3
= x² - 4x + 1

We get ,

x⁴ - 6x³ - 26x² + 138x - 35 = (x² - 4x + 1)(x² - 2x - 35)

The value of polynomial is also zero when ,

And, x + 5 = 0

or

x = -5

Hence, 7 and -5 are also zeroes of this polynomial.

To solve this question divide x⁴ - 6 x³ + 16 x² - 25 x + 10 by x² - 2 x + k by long division method
Let us divide, by

So, remainder = (2k - 9)x + (10 - 8k + k²)
But given remainder = x + a
⇒ (2k - 9)x + (10 - 8k + k²) = x + a
Comparing coefficient of x, we have
2k - 9 = 1
⇒ 2k = 10
⇒ k = 5
and
Comparing constant term,
10 - 8k + k² = a
⇒ a = 10 - 8(5) + 5²

⇒ a = 10 - 40 + 25
⇒ a = -5
So, the value of k is 5 and a is -5.