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i. p(x) = x² – 7x + 9 

Divisor = x + 1 

∴ take x = – 1 

∴ By remainder theorem, 

∴ Remainder =p(-1) 

p(x) = x² – 7x + 9 

∴ p(-1) = (- 1)² – 7(- 1) + 9 

= 1 + 7 + 9 

∴ Remainder = 17

ii. p(x) = 2x³ – 2x² + ax – a 

Divisor = x – a 

∴ take x = a 

By remainder theorem, 

Remainder = p(a) 

p(x) = 2x³ – 2x² + ax – a 

∴ p(a) = 2a³ – 2a² + a(a) – a 

= 2a³ – 2a² + a² – a 

∴ Remainder = 2a³ – a² – a 

iii. p(m) = 54m³ + 18m² – 27m + 5 

Divisor = m – 3 

∴ take m = 3 

∴ By remainder theorem, 

Remainder = p(3) 

p(m) = 54m³ + 18m² – 27m + 5 

∴ p(3) = 54(3)³ +18(3)² – 27(3) + 5 

= 54(27) + 18(9) – 81 + 5 

= 1458 + 162 – 81 + 5

∴ Remainder = 1544

Here, p(x) = x³ – 2mx + 21 

(x + 3) is a factor of x³ – 2mx + 21. 

∴ By factor theorem, 

Remainder = 0 

∴ P(- 3) = 0 

p(x) = x³ – 2mx + 21 

∴ p(-3) = (-3)³ – 2(m)(-3) + 21 

∴ 0 = – 27 + 6m + 21 

∴ 6 + 6m = 0 

∴ 6m = 6 

∴ m = 1 

∴ x + 3 is the factor of x³ – 2mx + 21 for m = 1.

Given polynomial is x⁴ – 16 

Let p(x) = x⁴ – 16 

We have p(-2) = (-2)⁴ – 16 

= 16 – 16 = 0 and 

p(2) = (2)⁴ – 16 

= 16 – 16 = 0 

p(-2) = 0 and p(2) = 0. 

So these are zeroes of the polynomial.

Given polynomial p(x) = x² – 9 

Zero of the polynomial p(x) = 0 

x² – 9 = 0 

⇒ x² = 9 

⇒ x = √9 = ± 3 

∴ x = + 3, – 3 

∴ Zeroes of the polynomial p(x) are – 3 and 3.

p(x) = 3x² – 4x + 7

p(1) = 3(1)² – 4(1) + 7 = 3 – 4 + 7 = 6

p(2) = 3(2)² - 4(2) + 7 = 12 – 8 + 7 = 11

p(3) = 3(3)² – 4(3) +7 = 27 – 12 + 7 = 22

p(l) + p(2) ≠ p(3)

p(6) = 3(6)² -4(6) + 7 = 108 – 24 + 7 = 91

P(2) × p(3) ≠ p(6)

Given polynomial p(x) = x² – x – 6 

We have, p(3) = 3² – 3 – 6 

= 9 – 3 – 6

= 9 – 9 

= 0 and 

p(-2) = (-2)² – (-2) – 6 

= 4 + 2 – 6 

= 6 – 6 

= 0 

We see that p(3) = 0 and p(-2) = 0 

∴ 3 and – 2 are the zeroes of the polynomial p(x).

The order of polynomial x² + 8x + 15 is 2. 

Hence the maximum number of zeroes of it is 2. 

Since the maximum possible zeroes of a polynomial is equal to its order.

If p(k) = 0 then ‘k’ is a zero value of p(x) 

Now p(x) = x² + 8x + 15 then 

p(6) = 6² + 8 (6) + 15 

= 36 + 48 + 15 ≠ 0 

As p(6) ≠ 0, ‘6’ cannot be the zero value of p(x).

Let us consider the given number = x Then its square = x²

8 times of it = 8x 

Now adding the above two x² + 8x result is = – 15 

x² + 8x = -15 

x² + 8x + 15 = 0 

x² + 5x + 3x + 15 = 0 

⇒ x(x + 5) + 3(x + 5) = 0 

⇒ (x + 3) (x + 5) = 0 

∴ x + 3 = 0 or x + 5 = 0 

Then x = – 3 or – 5

Correct option is (C) Not Real

\(\because\) \(x^2\geq0\)

\(\Rightarrow\) \(x^2+1\geq1\)

\(\Rightarrow\) \(x^2+1>0\)

Hence, \(x^2+1\) never gives a zero for any real number, but if \(x^2+1\) has a zero then x never be a real number.

Hence, if \(x^2+1\) has a zero then x is not a real number.

Correct option is  C) Not Real

p(x) = x² – 5x – 6 (given) . 

p(3) = 3² – 5(3) – 6 

= 9 – 15 – 6 

= 9 – 21 

= -12 

So p(3) = – 12.

We can write x² + 8x + 15 as x² + 5x + 3x + 15 

⇒ (x + 5) (x + 3) 

So its zero values are -5, -3

So the graph of x² + 8x + 15 intersect x axis at two points only.

A) Both A and B are true

Correct option is (D) n

The polynomial of degree n has n zeroes.

Correct option is  D) n

Correct option is (D) n

A polynomial of degree n will have n number of zeroes.

Correct option is D) n

Correct option is (A) 1

\(\frac{7x}{3}=-5\)

\(\Rightarrow\) \(x=-5\times\frac37\) \(=\frac{-15}7\)

\(\frac{-15}7\) is only zero of polynomial \(\frac{7x}{3}=-5\)

\(\therefore\) Number of zeros of polynomial \(\frac{7x}{3}=-5\) is 1.

Correct option is  A) 1

y = x² + 8x + 15 

= x² + 5x + 3x + 15 

= x (x + 5) + 3 (x + 5) 

y = (x + 3) (x + 5)

p(x) = x² – 5x + 6 (given) 

then p(3) = 3² – 5(3) + 6 

= 9 – 15 + 6 = 15 – 15 = 0 

∴ P(3) = 0

If the graph of the given quadratic polynomial touches X – axis at exactly one point, then I can confirm it has only one zero.

Correct option is (A) 2/3, 2/3

Let p(x) = \(9x^2-12x+4\)

p(x) = 0 gives \(9x^2-12x+4=0\)

\(\Rightarrow\) \((3x)^2-2\times3x\times2+2^2=0\)

\(\Rightarrow\) \((3x-2)^2=0\)

\(\Rightarrow\) 3x - 2 = 0 or 3x - 2 = 0

\(\Rightarrow\) x = \(\frac{2}{3}\) or x = \(\frac{2}{3}\)

\(\therefore\) \(\frac{2}{3}\;and\;\frac{2}{3}\) are zeroes of polynomial \(9x^2-12x+4.\)

Correct option is  A) 2/3 , 2/3

The quadratic polynomials y = 2x² – 4x + 5 and y = – 3x² + 2x – 1 and y = x² – 2x + 4 have no zeroes.

y = x² – x – 2 having two zeroes, i.e., (2, 0) and (- 1, 0). 

y = 3 – 2x – x² having two zeroes i.e., (1,0) and (- 3, 0). 

y = x² – 3x – 4 having two zeroes i.e., (-1, 0) and (4, 0)

(B) (c/a)

Explanation:

According to the question,

We have the polynomial,

ax³ + bx² + cx + d

We know that,

Sum of product of roots of a cubic equation is given by c/a

It is given that one root = 0

Now, let the other roots be α, β

So, we get,

αβ + β(0) + (0)α = c/a

αβ = c/a

Hence the product of other two roots is c/a

Hence, option (B) is the correct answer

i) Given polynomial is y = – x³ 

f(x) = -x³ ; f(x) = 0 ²

x³ = 0 

x = \(\sqrt[3]{0}\)= 0 

∴ Zero of the polynomial f(x) is only one i.e., 0.

ii) Given that y = x² – x³ 

f(x) = x² (1 – x) 

f(x) = 0 

⇒ x² (1 – x) = 0 

⇒ x² = 0 and 1 – x = 0 

⇒ x = 0 and x = 1 

∴ The zeroes of the polynomial f(x) are two i.e., 0 and 1.

iii) Given that x³ – 5x² + 6x Let f(x) = x³ – 5x² + 6x 

= x(x² – 5x + 6) 

= x(x² – 2x – 3x + 6) 

= x[x(x – 2) – 3(x – 2)] 

= x(x – 2) (x – 3) 

∴ The zeroes of the polynomial f(x) are x = 0 and x = 2 and x = 3

Let the zeroes of the quadratic polynomial be

α = √2, β = 2√2

Then, α + β = √2 + 2√2 = √2 (1 + 2) = 3√2

αβ = √2× 2√2 = 4

Sum of zeroes = α + β = 3√2

Product of zeroes = αβ = 4

Then, the quadratic polynomial

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

= x² – (3√2)x + 4

= x² – 3√2 x + 4

Let the zeros of the given polynomial be α, β and γ. (3 zeros as it’s a cubic polynomial) 

And given, the zeros are in A.P. 

So, let’s consider the roots as 

α = a – d, β = a and γ = a + d 

Where, a is the first term and d is the common difference. 

From given f(x), a = 2, b = -15, c = 37 and d = 30 

⇒ Sum of roots = α + β + γ 

= (a – d) + a + (a + d) 

= 3a 

= (\(\frac{-b}{a}\)) 

= -(\(\frac{-15}{2}\)) 

= \(\frac{15}{2}\) 

So, calculating for a, we get 3a = \(\frac{15}{2}\) 

⇒ a = \(\frac{5}{2}\) 

⇒ Product of roots = (a – d) x (a) x (a + d) 

= a(a² –d²) 

= \(\frac{-d}{a}\)

= \(\frac{-(30)}{2}\) 

= 15 

⇒ a(a² –d²) = 15 

Substituting the value of a, we get 

⇒ (\(\frac{5}{2}\))[(\(\frac{5}{2}\))² –d²] = 15 

⇒ 5[(\(\frac{25}{4}\)) –d²] = 30 

⇒ (\(\frac{25}{4}\)) – d² = 6 

⇒ 25 – 4d² = 24 

⇒ 1 = 4d² 

∴ d = \(\frac{1}{2}\) or -\(\frac{1}{2}\)

Taking d = \(\frac{1}{2}\) and a = \(\frac{5}{2}\) 

We get, 

the zeros as 2, \(\frac{5}{2}\) and 3 

Taking d = -\(\frac{1}{2}\) and a = \(\frac{5}{2}\) 

We get, the zeros as 3, \(\frac{5}{2}\) and 2.

Answer is (D) -21

Let f(x) = 2x² + x + k

One zero is 3

f(3) = 0

⇒ 2(3)² + 3 + k = 0

⇒ 2 × 9 + 3 + k = 0

⇒ 18 + 3 + k = 0

⇒ k = -21

Given quadratic polynomial

f(x) = x² + x – 2 = x² + 2x – x – 2 

= x (x + 2) – 1 (x + 2) = (x – 1) (x + 2)

To find zero, f(x) = 0

(x- 1) (x + 2) = 0

x – 1 = 0 or x + 2 = 0

x = 1 or x = -2

Thus 1 and -2 are two zeros of given polynomial

Relation between zeros and coefficient

Sum of zeros = 1 + (- 2) = – 1

and product of zeros = 1 × (-2) = -2

Comparing given polynomial with ax² + bx + c

a = 1, b = 1 and c = -2

Sum of zeros = -b/a = -1/1 = -1

and product of zeros = c/a = --2/1 = -2

Hence, relationship between zeros and coefficient is verified.

k[x² – (α + β)x + αβ]

No, x² – 1 cannot be the quotient on division of x⁶ + 2x³ + x – 1 by a polynomial in x of degree 5.

Justification:

When a degree 6 polynomial is divided by degree 5 polynomial,

The quotient will be of degree 1.

Assume that (x² – 1) divides the degree 6 polynomial with and the quotient obtained is degree 5 polynomial (1)

According to our assumption,

(degree 6 polynomial) = (x² – 1)(degree 5 polynomial) + r(x) [ Since, (a = bq + r)]

= (degree 7 polynomial) + r(x) [ Since, (x² term × x⁵ term = x⁷ term)]

= (degree 7 polynomial)

From the above equation, it is clear that, our assumption is contradicted.

x² – 1 cannot be the quotient on division of x⁶ + 2x³ + x – 1 by a polynomial in x of degree 5

Hence Proved.

(D) more than 3

Explanation:

According to the question,

The zeroes of the polynomials = -2 and 5

We know that the polynomial is of the form,

p(x) = ax² + bx + c.

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x² i.e.

Sum of the zeroes = – b/a

– 2 + 5 = – b/a

3 = – b/a

b = – 3 and a = 1

Product of the zeroes = constant term ÷ coefficient of x² i.e.

Product of zeroes = c/a

(- 2)5 = c/a

– 10 = c

Substituting the values of a, b and c in the polynomial p(x) = ax² + bx + c.

We get, x² – 3x – 10

Therefore, we can conclude that x can take any value.

Hence, option (D) is the correct answer.

The value of x for which polynomial f(x) = 0, is called zero of the polynomial.

L.C.M. × H.C.F. = f(x) × g(x)