Algebra Chapter 4 Polynomial
⇒ All algebraic expressions in which the indices of the variables are whole numbers are called Polynomials.
⇒ x = 1, x3 + 1, x2 – 9. x3 + x2 + 9…… all these are polynomials of which the variable is x, i.e. all these are polynomials with one variable. x2 + 8 is a Binomial.
⇒ 5x, x, 9x2 are called Monomials, and x3 – x2 + 2 is called a Trinomial. 0 is called zero polynomial.
⇒ Index of the highest power of the variable in any polynomial is called Degree of the polynomial. The degree of any constant polynomial is 0.
⇒ The degree of zero constant polynomial is indefined.
⇒ In case of determining the degree of a polynomial in more than one variable, we will find the sum of the indices of all variables of each term and the highest sum of indices will be the degree of that polynomial.
⇒ Degree of the polynomial f (x, y) = x5 + y2 + x3y3 + 4 is 3 + 3 = 6
⇒ A number C will be called zero of the polynomial f(x), if f(C) = 0.
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⇒ There is no zero of constant polynomial (non zero). Zero of zero constant polynomial cannot be defined.
⇒ Remainder Theorem: f(x) is a polynomial of degree x (x ≥ 1) and a is any real number.
⇒ If f (x) is divided by (x – a), then the remainder will be f(a).
⇒ Factor Theorem: If f (x) is any polynomial with degree x (x ≥ 1) and a is any real number, then
- (x – a) will be a factor of f (x), if (a) = 0
- f(a) = 0 if (x – a) is a factor of f (x).
Algebra Chapter 4 Polynomial Fill In The Blanks
Example 1. The co-efficient of the term x of the polynomial 10 is _______
Solution: 0.
Example 2. The polynomials which have only one term are called _______
Solution: monomial.
Example 3. The polynomials having four terms are called _______
Solution: tetramonials.
Example 4. Polynomials of degree 0 are called ______ polynomials.
Solution: constant.
Example 5. The co-efficient of x0 of the polynomial √11- 3√11x + x2 is ______
Solution: √11
Example 6. x + \(\frac{5}{x}\) is _______ a polynomial.
Solution: not.
Example 7. Polynomial having degree 4 are called _______ polynomial.
Solution: bignadratic
Example 8. The polynomial of degree 3 are called _______ polynomial.
Solution: cubic.
Example 9. bx + c is a ______ polynomial.
Solution: linear.
Example 10. Graph of any linear polynomial is a _______ line.
Solution: straight.
Algebra Chapter 4 Polynomial True Or False
Example 1. A binomial can have at most two terms.
Solution: The statement is False.
Example 2. Every polynomial is a binomial.
Solution: The statement is False.
Example 3. A binomial may have degree 5.
Solution: The statement is True.
Example 4. Zero of a polynomial is always 0.
Solution: The statement is False.
Example 5. A polynomial cannot have more than one zero.
Solution: The statement is False.
Example 6. The degree of the sum of two polynomials cach of degree 5 is always 5.
Solution: The statement is True.
Example 7. √5x is a linear polynomial.
Solution: The statement is False.
Example 8. The division between two polynomials may or may not be a polynomial.
Solution: The statement is True.
Example 9. The sum of two polynomials is always a polynomial.
Solution: The statement is True.
Example 10. The difference of two polynomials may or may not be a polynomial.
Solution: The statement is False.
Algebra Chapter 4 Polynomial Short Answer Type Questions
Example 1. Let us write the zero of the polynomial P(x) = 2x – 3
Solution: 2x – 3 = 0, x = \(\frac{3}{2}\)
Example 2. If P(x) = x + 4, let us write the value of P(x) + P(-x)
Solution: P(x) + P(-x) = x + 4x +4 = 8
Example 3. Let us write the remainder if the polynomial x3 + 4x2 + 4x – 3 is divided by x
Solution: Zero of polynomial x is 0
Remainder = f(0) = 03 + 402 + 4.0 – 3 (by Remainder theorem)
= -3
Example 4. If (3x – 1)7 = a7x7 + a6x6 + a5x5 +…….. +a1x + a0 then find the value of a7 + a6 + a5 + ……….+ a0.
Solution: It is an identify, we put x = 1
∴ (3 × 1 – 1)7 = a7(1)7 + a6(1)6 + a5(1)5 +…….a1.1 + a0
⇒ or, 128 = a7 + a6 + a5 + …….. + a1 + a0
Example 5. For the polynomial \(\frac{x^3+2 x+1}{5}-\frac{7}{2} x^2-x^6\), find
- The degree of the polynomial
- The co-efficient of x3
- The co-efficient of x6
- The coefficient of x0
Solution:
- degree is 6,
- \(\frac{1}{5}\)
- -1
- \(\frac{1}{5}\)
Example 6. If f(x)=3x3 -4x2 + 7x – 5, find f(3), f (-3)
Solution: f(3) = 3.33. 4.32 + 7.3 – 5
= 81 – 36 + 21 – 5 = 102 – 41 = 61.
f(-3) = 3(-3)3 – 4(-3)2 + 7(-3) – 5 = -81 – 36 – 21 – 5 = -143
Example 7. Calculate and write the value of a for which (x + a) will be a factor of the polynomial x3 + ax2 – 2x + a – 12.
Solution: Let f (a) = x3 + ax2 – 2x + a – 12
⇒ We have to find the zero of the linear polynomial x + a
⇒ x + a = 0, x = -a
∴ by factor theorem f(-a) = 0
⇒ or, (-a)3 + a (-a)2 – 2(-a) + a – 12 = 0
⇒ or, (-a)3 + a3 + 2a + a – 12 = 0
⇒ or, 3a = 12
∴ a = 4
Example 8. Find the value of k for which (x-3) will be a factor of the polynomial k2x3 – kx2 + 3kx – k
Solution: First find the zero of the linear polynomial (x-3),
⇒ x – 3= 0, x = 3
⇒ By factor theorem f(3) = 0
⇒ or, k2(3)3 – k (3)2 + 3k.3 – k = 0
⇒ or, 27k2 – 9k + 9k -k = 0
⇒ or, k (27k – 1) = 0
∴ k =0, \(\frac{1}{27}\)
Example 9. Let us write the value of f (x) + f (-x) when f(x) = 2x + 5.
Solution: f(x) + f (-x) = 2 + 5 + 2 (-x)+ 5 = 10
Example 10. Both (x – 2) and (x – \(\frac{1}{2}\)) are factors of the polynomial px2 + 5x + r, let us calculate and write the relation between p and r.
Solution: Zeros of the polynomials (x-2) and (x – \(\frac{1}{2}\))
⇒ x – 2 = 0 x = 2 and x – \(\frac{1}{2}\) = 0 x = \(\frac{1}{2}\)
⇒ Now, p(2)2 + 5(2) + r = 0
⇒ or, 4p + r = -10….. (1)
⇒ and p\(\left(\frac{1}{2}\right)^2\) + 5\(\frac{1}{2}\) + r = 0
⇒ or, \(\frac{p}{4}+\frac{5}{2}+r=0\)
⇒ or, \(\frac{p}{4}+r=-\frac{5}{2}\)……..(2)
⇒ By (1) – (2), we have,
⇒ \(4 p-\frac{p}{4}=-10+\frac{5}{2}\)
⇒ \(\frac{18 p}{4_2}=\frac{-18}{2}\)
⇒ or, p = -2
∴ p = r.
⇒ r = – 10 – 4p
⇒ = -10 – 4(-2) = -2
Example 11. Find the roots of the polynomial f(x)= 2x + 3
Solution: Zeros of the polynomials are the roots.
⇒ To find the zero of linear polynomial 2x + 3
⇒ 2x + 3 = 0; x = –\(\frac{3}{2}\)
Example 12. Check the following statement, ‘The two zeros of the polynomial p(x)= x2 – 9 are 3, -3′
Solution: x2 – 9 = 0,
⇒ x = ±3
⇒ The given statement is true.
Example 13. Find the number of terms of the polynomial \(\left(\frac{x+2 x^2+x^3}{x}\right)^n\)
Solution: \(\left(\frac{x+2 x^2+x^3}{x}\right)^n=\left\{\frac{x\left(1+2 x+x^2\right)}{x}\right\}^n=\left(1+2 x+x^2\right)^n\)
= (1 + x)2x number of terms = 2x + 1
Example 14. If f(x) = 2ax + 1, then find the value of f(a).f(b).f(c)
Solution: f(a).f(b).f(c) = 2a.a+1.2ab+1 .2ac+1
= 2a2 + 1 + ab + 1 + ac + 1
= 2a(a + b+ c) + 1.22 = 4.2ax+1
When x = ax + b + c
= 4f(a + b + c)
Example 15. If f(x) = 2x, then show that f(x + 1) = 4f(x – 1)
Solution: f(x + 1) = 2x+1 = 2.2x
4f(x – 1) = 4.2x-1 = \(\frac{4.2^x}{2}\) = 2.2x
∴ f(x + 1) = 4 + (x – 1)