Question

Class 9th
Mathematics
Chapter 2 - Polynomials
Formulas Needed . Give 1 example for each Formuals

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Answer 1

EXPLANATION.

Polynomial.

(1) = Relationship between zeroes and coefficient of a polynomial.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

Formula of the quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Example :

Find the zeroes of the quadratic polynomial x² + 7x + 10, and verify the relationship between the zeroes and the coefficients.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -/a.

⇒ α + β = -(7)/1 = -7.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 10/1 = 10.

As we know that,

Factorizes the equation into middle term splits, we get.

⇒ x² + 7x + 10 = 0.

⇒ x² + 5x + 2x + 10 = 0.

⇒ x(x + 5) + 2(x + 5) = 0.

⇒ (x + 2)(x + 5) = 0.

⇒ x = -2  and  x = -5.

Sum of the value of x.

⇒ - 2 + (-5) = -7.

Products of the value of x.

⇒ (-2)(-5) = 10.

Hence verified.

Example :

Find a quadratic polynomial, the sum and the products of whose zeroes are  -3 and 2 respectively.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -3.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 2.

Formula of the quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (-3)x + 2 = 0.

⇒ x² + 3x + 2 = 0.

(2) = Relationship between zeroes and coefficients.

As we know that,

Sum of the zeroes of the cubic polynomial.

⇒ α + β + γ = -b/a.

Products of the zeroes of a cubic polynomial two at a time.

⇒ αβ + βγ + γα = c/a.

Products of the zeroes of the cubic polynomial.

⇒ αβγ = -d/a.

Example :

Verify that 3, -1, -1/3 are the zeroes of the cubic polynomial p(x) = 3x³ - 5x² - 11x - 3, and then verify the relationship between the zeroes and the coefficients.

As we know that,

Sum of the zeroes of the cubic polynomial.

⇒ α + β + γ = -b/a.

⇒ α + β + γ = -(-5)/3 = 5/3.

Products of the zeroes of the cubic polynomial two at a time.

⇒ αβ + βγ + γα = c/a.

⇒ αβ + βγ + γα = -11/3.

Products of the zeroes of the cubic polynomial.

⇒ αβγ = -d/a.

⇒ αβγ = -3/3 = -1.

As we know that,

⇒ p(x) = 3x³ - 5x² - 11x - 3

Put the value of x = 3 in equation, we get.

⇒ p(3) = 3(3)³ - 5(3)² - 11(3) - 3.

⇒ p(3) = 81 - 45 - 33 - 3.

⇒ p(3) = 81 - 81.

⇒ p(3) = 0.

Put the value of x = -1 in equation, we get.

⇒ p(-1) = 3(-1)³ - 5(-1)² - 11(-1) - 3.

⇒ p(-1) = -3 - 5 + 11 - 3.

⇒ p(-1) = 11 - 11.

⇒ p(-1) = 0.

Put the value of x = -1/3 in equation, we get.

⇒ p(-1/3) = 3(-1/3)³ - 5(-1/3)² - 11(-1/3) - 3.

⇒ p(-1/3) = -1/9 - 5/9 + 11/3 - 3.

⇒ p(-1/3) = - 1 - 5 + 33 - 27/9.

⇒ p(-1/3) = 33 - 33/9.

⇒ p(-1/3) = 0.

Hence verified.

(3) = Division Algorithm for polynomials.

If p(x) and g(x) are any two polynomial with g(x) ≠ 0, then we find polynomial q(x) and r(x) such that,

p(x) = g(x) x q(x) + r(x).

Example :

Find all zeroes of 2x⁴ - 3x³ - 3x² + 6x - 2, if you know that two of its zeroes are √2 and -√2.

As we know that,

Zeroes of the polynomial,

⇒ x = √2.

⇒ x - √2.

⇒ x = -√2.

⇒ x + √2.

Products of the zeroes of the quadratic equation.

⇒ (x - √2)(x + √2).

As we know that,

Formula of :

⇒ (x² - y²) = (x - y)(x + y).

Using this formula in equation, we get.

⇒ x² - 2.

Divide 2x⁴ - 3x³ - 3x² + 6x - 2 by x² - 2 we get,

⇒ 2x² - 3x + 1.

Factorizes the equation into middle term splits, we get.

⇒ 2x² - 2x - x + 1 = 0.

⇒ 2x(x - 1) - 1(x - 1) = 0.

⇒ (x - 1)(2x - 1) = 0.

⇒ x = 1  and  x = 1/2.

Zeroes of the polynomial = √2 , -√2 , 1/2 , 1.

Answer 2

$\large \tt \: Some \: of \: the \: important \: formulas / theroms :-$

Degree of Polynomial:-

Highest power of variable in a polynomial is called degree of polynomial.

E.g. $\tt \: p(x)=9x {}^{3} - 3x {}^{2} +8x – 2,$

the highest power is 3.

Classification of Polynomials based on degree

• Linear Polynomial: Degree is 1.
• Quadratic Polynomial: Degree is 2
• Cubic Polynomial: Degree is 3.

General form of polynomials of various degrees.

• $\tt \: Linear \: Polynomial = p(x) = ax + b , where \: a≠ 0$

• $\tt \: Quadratic \: Polynomial = p(y) = a {x}^{2} + by + c , \: where a≠ 0$

• $\tt \: Cubic \: Polynomial = p(x) = ax {}^{3} + bx {}^{2} + cx + d , where \: a≠ 0$

Zeroes of Polynomial

Zeroes of a polynomial p(x) is real number ‘a’ for which polynomial p(x) if p(a) = 0.

E.g.: For p(x) = x-2 ,

p(2) = 2-2 =0.

Thus 2 is zeroes for polynomial p(x)= x-2

Zeroes of a polynomial is special. It is used to find factors of equation p(x)=0.

Remainder theorem:

Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).

Question: Find the remainder when $\tt \: p(x) = x {}^{4} + x {}^{3} – 2x {}^{2} + x + 1$ is divided by x – 1.

Solution: Zero of x – 1 is 1, so as per remainder theorem remainder in this case will be p(1) .

$\tt \: So, p(1) = (1) {}^{4} + (1) {}^{3} – 2(1) {}^{2} + 1 + 1 = 2$

Factor Therom:

x – a is a factor of the polynomial p(x), if p(a) = 0. Also, if x – a is a factor of p(x), then p(a) = 0, where a is any real number. This is an extension to remainder theorem where remainder is 0, i.e. p(a) = 0.

Numerical: Examine whether x + 2 is a factor of

$\tt \: p(x)= x {}^{3} + 3x {}^{2} + 5x + 6$

Solution : The zero of x + 2 is –2. As, per factor theorem, x+2 is factor of p(x) if p(-2) = 0.

$\tt \: p(–2) = (–2) {}^{3} + 3(–2) {}^{2} + 5(–2) + 6 = 0.$

Thus, x+2 is factor of $\tt \: p(x)= x {}^{3} + 3x {}^{2} + 5x + 6$

Algebraic identity

is an algebraic equation that is true for all values of the variables occurring in it

$\tt \: (x + y) {}^{2} = x {}^{2} + 2xy + y {}^{2} \\ \tt \: (x – y) {}^{2} = x {}^{2} – 2xy + y {}^{2} \\ \tt \: x{}^{2} – y {}^{2} = (x + y) (x – y)\\ \tt \: (x + a) (x + b) = x {}^{2} + (a + b)x + ab \\ \tt \: (x + y + z) {}^{2} = x {}^{2} + y {}^{2} + z {}^{2} + 2xy + 2yz + 2zx\\ \tt \: (x + y) {}^{ 3} = x {}^{3} + y {}^{3} + 3xy(x + y)\\ \tt \: (x – y) {}^{3} = x {}^{3} – y {}^{3} – 3xy(x – y)\\ \tt \: x {}^{3} + y{}^{3}+ z{}^{3}– 3xyz = (x + y + z) (x {}^{2} + y {}^{2} + z {}^{2} – xy – yz – zx)$

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