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p(x) = x⁴ – 2x² – x 

∴ p(1) = (1)⁴ – 2(1)² – 1 

= 1 – 2 – 1 

∴ p(1) = -2 

∴ p(x) = x⁴ – 2x² – x

 ∴ p(0) = (0)⁴ – 2(0)² – 0 

= 0 – 0 – 0 

∴ p(0) = 0

p(x) = x⁴ – 2x² – x 

∴ p(-2) = (-2)⁴ – 2(-2)² – (-2)

 = 16 – 2(4) + 2 

= 16 – 8 + 2 

∴ p(-2) = 10

p(m) = m³ + 2m + a 

∴ p(2) = (2)³ + 2(2) + a 

∴ 12 = 8 + 4 + a … [∵ p(2)= 12] 

∴ 12 = 12 + a 

∴ a = 12 – 12 

∴ a = 0

p(x) = mx² – 2x + 3 

∴ p(- 1) = m (- 1)² – 2(- 1) + 3 

∴ 7 = m(1) + 2 + 3 …[∵ p(-1) = 7] 

∴ 7 = m + 5 

∴ m = 7 – 5 

∴ m = 2

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 + a – a 

∴ Remainder = 2a³ – a² – a

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

Statement:

If p(x), a polynomial in x is divided by x-a and the remainder = p (a) is zero, then (x-a) is a factor of p(x).

Proof:

When p(x) is divided by x-a,

R = p(a) (by remainder theorem)

p(x) = (x-a).q(x)+p(a)

(Dividend = Divisor x quotient + Remainder Division Algorithm)

But p(a) = 0 is given.

Hence p(x) = (x-a).q(x)

=> (x-a) is a factor of p(x).

Conversely if x-a is a factor of p(x) then p(a) = 0.

p(x) = (x-a).q(x) + R

If (x-a) is a factor, then the remainder should be zero (x - a divides p(x)

exactly)

R = 0

By remainder theorem, R = p(a)

=> p(a) = 0

If f(x) is a polynomial in x and is divided by x-a; the remainder is the value of f(x) at x = a i.e., Remainder = f(a).

Proof:

Let p(x) be a polynomial divided by (x-a).

Let q(x) be the quotient and R be the remainder.

By division algorithm,

Dividend = (Divisor x quotient) + Remainder

p(x) = q(x) . (x-a) + R

Substitute x = a,

p(a) = q(a) (a-a) + R

p(a) = R (a - a = 0, 0 - q (a) = 0)

Hence Remainder = p(a).

Steps for Factorization using Remainder Theorem

• By trial and error method, find the factor of the constant for which the given expression becomes equal to zero.
• Divide the expression by the factor that is determined in step 1.
• Factorize the quotient. If the quotient is a trinomial, factorize it further.
• If the expression is a 4^th degree expression, the first step will be to reduce this to a trinomial and then factorize this trinomial further.

• Find all pairs of factors whose product is the last term of the trinomial.
• From the pairs of factors from step 1, choose a pair of factors whose sum is the coefficient of the middle term of the trinomial.
• Split the middle term using the pair of factors from step 2 and rewrite the trinomial.
• Group the terms from step 3 and factorize.
• Verify the solution.

Factoring a trinomial of the type ax+bx+bx²+c(a≠1) by splitting the middle term

To factorize expressions of the type x² + bx + c, you will find two numbers a and b such that their sum is equal to the coefficient of the middle term and their product is equal to the last term(constant).

Steps for factoring ax²+bx+c(a≠1) by grouping

• Find the product (ac), of the coefficient of x² and the last term.
• List the factor pairs of ac.
• Choose a factor pair whose sum is the coefficient of the middle term.
• Rewrite the polynomial by splitting the middle term.
• Regroup and factorize.

In many situations, we come across polynomials, which may not have common factors among its terms. In such cases, we group the terms of the expression in such a way that there are common factors among the terms of the groups so formed.

Steps for Factorization by Grouping

• Rearrange the terms if necessary.
• Group the given expression in such a way that each group has its common factor.
• Identify the HCF of each group.
• Identify the other factor.
• Write the expression as a product of the common factor and the other factor.

HCF of a polynomial is the largest monomial, which is a factor of each term of the polynomial.

We can factorize a polynomial by finding the Highest Common Factor (HCF) of the terms of the expression and then dividing each term by its HCF. HCF and the quotient obtained are the factors of the given expression.

Steps for Factorization

• Identify the HCF of the terms of the given expression.
• Divide each term of the given expression by the HCF and find the quotient.
• Write the given expression as a product of HCF and quotient.

There are various methods of factorizing a polynomial. They are,

1. Factorization by dividing the expression by the HCF of the terms of the given expression.

2. Factorization by grouping the terms of the expression.

3. Factorization using identities.

As the graph intersects x-axis at one Point

.'. The number of zeroes of p(x) is 1

 f(x) = x² - 7x - 8

Let other zero be k, then

Sum of zeroes -1 + k = -(-7/1) = 7

p(x) = 3x² - kx + 6

sum of the zeroes = 3 = -(Cofficient of x)/(Cofficient of x²)

3 = -((-k)/3)

k = 9

Let    P(x) = 4x² -12x + 9

=  4x² - 6x + 9

or,  p(x)  =  2x(2x - 3)  -3(2x  - 3)

 :.    0  =  (2x - 3) (2x - 3)

The zeros are    3 / 2, 3 / 2

:.    x = 3 / 2, 3 / 2

Hence, zeroes of the polynomial  are 3 / 2, 3 / 2.

Let f(x) = 2x³ + ax² + 3x – 5 

g(x) = x³ + x² – 4x + a 

Given that f(x) and g(x) divided x – 2 give same remainder. 

i e., f(2) = g(2) 

By Remainder theorem. 

But f(2) = 2(2)³ + a(2) + 3(2)² – 5 

= 2 x 8 + 4a + 6 – 5 

= 17 +4a 

g(2) = 2³ + 2² – 4(2) + a. 

= 8 + 4 – 8 + a = 4 + a 

i.e., 4 + a = 17 + 4a 

∴ a – 4a = 17 – 4 – 3a = 13 

a = -13/3

Correct option is (B) 3abc

If a + b + c = 0, then \(a^3+b^3+c^3\) = 3abc

Correct option is  B) 3abc

Correct option is (C) (x – 2) (x² + 2x + 4)

\(x^3-8=x^3-2^3\) \(=(x-2)\,(x^2+2x+2^2)\)

\(=(x-2)\,(x^2+2x+4)\)

C) (x – 2) (x² + 2x + 4) 

Correct option is (D) f(a)

If \(f(x)\) is divided by \((x-a)\), then the remainder is f(a).

Correct option is  D) f(a)

Correct option is (B) 3x – 2

\(3x^2+x-2=3x^2+3x-2x-2\)

\(=3x(x+1)-2(x+1)\)

\(=(x+1)(3x-2)\)

Hence \((x+1)\) and \((3x-2)\) are factors of polynomial \((3x^2+x-2).\)

Correct option is  B) 3x – 2

Correct option is (A) 2(x² + y²)

\((x+y)^2+(x-y)^2\) \(=(x^2+y^2+2xy)+(x^2+y^2-2xy)\)

= \(2(x^2+y^2)\)

A) 2(x² + y² ) 

Correct option is (C) (3x + 5) (3x – 5)

\(9x^2-25=(3x)^2-5^2\)

= \((3x+5)\) \((3x-5)\)

C) (3x + 5) (3x – 5)

Correct option is (A) (x – 2) (x + 2) – (x² + 5x)

\((x – 2) (x + 2) – (x^2 + 5x)\)

\(=x^2-4-x^2-5x\)

= -4 - 5x which is a linear polynomial not a quadratic polynomial.

Correct option is A) (x – 2) (x + 2) – (x² + 5x)

Let a,6 be the zeros of given polynomial.

Then a+6=5 ⇒ a=-1.

We know that,

p(x)= g(x) × q(x)+r(x)

According to the question,

q(x) =0

When q(x)=0, then r(x) is also = 0

So, now when we divide p(x) by g(x),

Then p(x) should be equal to zero

Hence, the relation between the degrees of p (x) and g (x) is the degree p(x)<degree g(x)

sum of zeros =-3+4=1,

product of  zeros =-3x4=-12

Required polynomial =x²-x-12

Product ac = 24 & b = -14 

∴ Split the middle term as - 12 & - 2 

⇒ x² - 14x + 24 

= x² - 12 - 2x + 24 

⇒ x(x - 12) - 2 (x - 12) = (x - 12)(x - 2) 

2x² + 12 √2x + 35 

Product ac = 70 & b = 12 √2 

∴ Split the middle term as 7 √2 &5 √2

⇒ 2x² + 12 √2x + 35 = 2x² + 7 √2x + 5 √2x + 35 

 = √2x[ √2x + 7]+ 5[ √2x + 7] 

 = [ √2x + 5][ √2x + 7]

45 ⇒ ±1,±3,±5,±9,±15,±45

if we put x = 1 in p(x) 

 p(1) = 2(1)⁴ - 7(1)³ - 13(1)² + 63(1) - 45 

 2 - 7 - 13 + 63 - 45 = 65 - 65 = 0 

∴ x = 1 or x - 1 is a factor of p(x).

Similarly, if we put x = 3 in p(x) 

 p(3) = 2(3)⁴ - 7(3)³ - 13(3)² + 63(3) - 45 

 162 - 189 - 117 + 189 - 45 = 162 - 162 = 0

Hence, x = 3 or x - 3 = 0 is the factor of p(x). 

 p(x) = 2x⁴ - 7x³ - 13x² + 63x - 45 

∴ p(x) = 2x³ (x - 1) -5x² (x - 1) - 18(x - 1) + 45(x - 1) 

 2x⁴ - 2x³ (x - 1) - 5x² - 18x² + 18x + 45x - 54 

⇒ p(x) = (x - 1)(2x³ - 5x² - 18x + 45) 

⇒ p(x) = (x - 1)(2x³ - 5x² - 18x + 45) 

⇒ p(x) = (x - 1)[2x² (x - 3) + x(x - 3) - 15(x - 3)] 

⇒ p(x) = (x - 1)[2x³ - 6x² + x² - 3x - 15x + 45] 

⇒ p(x) = (x - 1)(x - 3)(2x² + x - 15) 

⇒ p(x) = (x - 1)(x - 3)(2x² + 6x - 5x - 15) 

⇒ p(x) = (x - 1)(x - 3)[2x(x + 3) - 5(x + 3)] 

⇒ p(x) = (x - 1)(x - 3)(x + 3)(2x - 5)

General form of a quadratic polynomial having variable ‘x’ is 

f(x) = ax³ + bx² + c, a ≠ 0 

General form of a cubic polynomial having variable ‘x’ is 

f(x) = ax³ + bx² + cx + d, a ≠ 0

(i) p(x) = x – 4

Zero of the polynomial p(x) ⇒ p(x) = 0

P(x) = 0

⇒ x – 4= 0

⇒ x = 4

Therefore, the zero of the polynomial is 4.

(ii) g(x) = 3 – 6x

Zero of the polynomial g(x) ⇒ g(x) = 0

g(x) = 0

⇒3 – 6x = 0

⇒ x = 3/6 = ½

Therefore, the zero of the polynomial is ½

(iii) q(x) = 2x –7

Zero of the polynomial q(x) ⇒ q(x) = 0

q(x) = 0

⇒2x – 7 = 0

⇒ x = 7/2

Therefore, the zero of the polynomial is 7/2

(iv) h(y) = 2y

Zero of the polynomial h(y) ⇒ h(y) = 0

h(y) = 0

⇒2y =0

⇒ y = 0

Therefore, the zero of the polynomial is 0

(i) –3 is a zero of x – 3

False

Zero of x – 3 is given by,

x – 3 = 0

⇒ x=3

(ii) – 1/3 is a zero of 3x + 1

True

Zero of 3x + 1 is given by,

3x + 1 = 0

⇒ 3x = – 1

⇒ x = – 1/3

(iii) – 4/5 is a zero of 4 –5y

False

Zero of 4 – 5y is given by,

4 – 5y =0

⇒ – 5y = – 4

⇒ y = 4/5

(iv) 0 and 2 are the zeroes of t² – 2t

True

Zeros of t² – 2t is given by,

t² – 2t = t(t – 2) = 0

⇒ t = 0 or 2

(v) –3 is a zero of y² + y – 6

True

Zero of y² + y – 6 is given by,

y² + y – 6 = 0

⇒ y² + 3x – 2x – 6 = 0

⇒ y (y + 3) – 2(x + 3) = 0

⇒ (y – 2) (y + 3) =0

⇒ y = 2 or – 3