# Worksheets For Class 9 Mathematics Polynomials

Polynomials Class 9 Worksheet PDF have been designed as per the latest pattern for CBSE, NCERT and KVS for Grade 9. Students are always suggested to solve printable worksheets for Mathematics Polynomials Grade 9 as they can be really helpful to clear their concepts and improve problem solving skills. We at worksheetsbag.com have provided here free PDF worksheets for students in standard 9 so that you can easily take print of these test sheets and use them daily for practice. All worksheets are easy to download and have been designed by teachers of Class 9 for benefit of students and is available for free download.

## Mathematics Polynomials Worksheets for Class 9

We have provided chapter-wise worksheets for class 9  Mathematics Polynomials which the students can download in Pdf format for free. This is the best collection of  Mathematics Polynomials standard 9th worksheets with important questions and answers for each grade 9th  Mathematics Polynomials chapter so that the students are able to properly practice and gain more marks in Class 9  Mathematics Polynomials class tests and exams.

1. Polynomials in one Variable
2. Zeroes of a Polynomial
3. Remainder Theorem
4. Factorisation of Polynomials
5. Algebraic Identities

• Constants: A symbol having a fixed numerical value is called a constant.
• Variables: A symbol which may be assigned different numerical values is known as variable.
• Algebraic expressions: A combination of constants and variables. Connected by some or all of the operations +, -, X and is known as algebraic expression.
• Terms: The several parts of an algebraic expression separated by ‘+’ or ‘-‘ operations are called the terms of the expression.
• Polynomials: An algebraic expression in which the variables involved have only nonnegative integral powers is called a polynomial.
(i) 5x2 – 4x2 – 6x – 3 is a polynomial in variable x.
(ii) 5 + 8x3/2 + 4x-2 is an expression but not a polynomial.
• Coefficients: In the polynomial x3 + 3x2 + 3x 1+, coefficient of x3, x2 , x are1,3, 3 respectively and we also say that +1 is the constant term in it.
• Degree of a polynomial in one variable: In case of a polynomial in one variable the highest power of the variable is called the degree of the polynomial.
• Classification of polynomials on the basis of degree.
Degree                Polynomial            Example
(a) 1                   Linear                     x +1, 2x + 3etc.
(b) 2                   Quadratic              ax2 + bx + c etc.
(c) 3                   Cubic                     x3 + 3x2 +1 etc. etc.
Classification of polynomials on the basis of no. of terms
No. of terms Polynomial & Examples.
(i) 1 Monomial – 1/3
(ii) 2 Binomial – (3+ 6x), (x – 5y) etc.
(iii) 3 Trinomial- 2×2 + 4x + 2 etc. etc.
· Constant polynomial: A polynomial containing one term only, consisting a constant term is called a constant polynomial the degree of non-zero constant polynomial is zero.
· Zero polynomial: A polynomial consisting of one term, namely zero only is called a zero polynomial. The degree of zero polynomial is not defined.
· Zeroes of a polynomial: Let p(x) be a polynomial. If p(a) =0, then we say that is a zero of the polynomial of p(x).
· Remark: Finding the zeroes of polynomial p(x) means solving the equation p(x)=0.
· Remainder theorem: Let f (x) be a polynomial of degree ≥ 1 and let a be any real number. When f(x) is divided by (x – a) then the remainder is f (a)
· Factor theorem: Let f(x) be a polynomial of degree n > 1 and let a be any real number.
(i) If f (a) = 0 then (x – a) is factor of f (x)
(ii) If (x – a) is factor of f (x)then f (a) = 0
· Factor: A polynomial p(x) is called factor of q(x) divides q(x) exactly.
· Factorization: To express a given polynomial as the product of polynomials each of degree less than that of the given polynomial such that no such a factor has a factor of lower degree, is called factorization.

1. Which of the following is a polynomials?

2. √2 is a polynomial of degree
(a) 2
(b) 0
(c) 1
(d) 1/2

3. Degree of the polynomial of 4 3 5 4x + 0x + 0x + 5x + 7 is
(a) 4
(b) 5
(c) 3
(d) 7

4. Degree of the zero polynomial
(a) 0
(b) 1
(c) Any natural number
(d) Not defined.

5. If 2 p(x) = x − 2√2x +1, then p(2√2) is equal to
(a) 0
(b) 1
(c) 4√2
(d) 8√2 +1

6. The value of the polynomial 5x − 4x2 + 3, when x = −1 is
(a) – 6
(b) 6
(c) 2
(d) – 2

7. If p(x) = x + 3, then p(x) + p(−x) is equal to
(a) 3
(b) 2x
(c) 0
(d) 6

8. Zero of the zero polynomial is
(a) 0
(b) 1
(c) Any real number
(d) Not defined

9. Zero of the polynomial p(x) = 2x + 5 is
(a) -(2/5)
(b) -(5/2)
(c) 2/5
(d) 5/2

10. One of the zeroes of the polynomial 2x2 + 7x − 4 is
(a) 2
(b) 1/2
(c) -(1/2)
(d) −2

11. If x51 + 51 is divided by x + 1, the remainder is
(a) 0
(b) 1
(c) 49
(d) 50

12. If x +1, is a factor of the polynomial 2x2 + kx, then the value of k is
(a) – 3
(b) 4
(c) 2
(d) – 2

13. x + 1, is a factor of the polynomial
(a) x3 + x2 − x +1
(b) x3 + x2 + x +1
(c) x4 + x3 + x2 +1
(d) x4 + 3x3 + 3x2 + x +1

14. One of the factor of (25x2 −1) + (1+ 5x2) is
(a) 5 + x
(b) 5 – x
(c) 5x – 1
(d) 10x

15. The value of 2492 − 2482 is
(a) 12
(b) 477
(c) 487
(d) 497

16. The factorization of = 4x2 +8x + 3 is
(a) (x +1) (x + 3)
(b) (2x +1) (2x + 3)
(c) (2x + 2) (2x +5)
(d) (2x −1) (2x −3)

17. Which of the following is a factor of (x + y)2 − (x3 + y3 ) ?
(a) x2 + y2 + 2xy
(b) x2 + y2 − xy
(c) xy2
(d) 3xy

18. The coefficient of x in the expansion of (x + 3)3 is
(a) 1
(b) 9
(c) 18
(d) 27

19. If x/y + y/x = − 1 the value of x3 − y3 is
(a) 1
(b) – 1
(c) 0
(d) 1/2

20. If 49 x2 – b =

(a) 0
(b) 1/√2
(c) 1/4
(d) 1/2

21. If a + b + c = 0, then the value of a + b + c is equal to
(a) 0
(b) abc
(c) 3abc
(d) 2abc

Sol. (i) 8 is a constant polynomial.
(ii) √3x2 − 2x
In each term of this expression, the exponent of the variable x is a whole number. Hence, it is a polynomial.
(iii)1− √5x = 1− √5x1/2
Here, the exponent of the second term, i.e.,x1/2, 1/2, which is not a whole number Hence, the given algebraic expression is not a polynomial.
(iv) 1/5x-2 + 5x + 7 = (1/5) x2 + 5x + 7
In each term of this expression, the exponent of the variable x is a whole number. Hence, it is a polynomial.
(v) (x – 2) (x – 4) x = x2 – 6x + 8 / x = x – 6 + 8 / x = x – 6 + 8x-1
Here, the exponent of variable x in the third term, i.e., in 8x −1 is –1, which is not a whole number. So, this algebraic expression is not a polynomial.
(vi) 1 / x + 1 = (x + 1)−1 which cannot be reduced to an expression in which the exponent of the variable x have only whole numbers in each of its terms. So, this algebraic expression is not a polynomial.

In this expression, the exponent of a in each term is a whole number, so this expression is a polynomial.

(viii) 1/2x = (1/2) x-1
Here, the exponent of the variable x is – 1, which is not a whole number so, this algebraic expression is not a polynomial.

2. Write whether the following statements are True or False. Justify your answer.
(i) A binomial can have at most two terms
(ii) Every polynomial is a binomial
(iii) A binomial may have degree 5
(iv) Zero of a polynomial is always 0
(v) A polynomial cannot have more than one zero
(vi) The degree of the sum of two polynomials each of degree 5 is always 5.
Sol. (i) The given statement is false because binomial have exactly two terms.
(ii) A polynomial can be a monomial, binomial trinomial or can have finite number of terms. For example, x4 + x3 + x2 +1 is a polynomial but not binomial.
Hence, the given statement is false.
(iii) The given statement is true because a binomial is a polynomial whose degree is a whole number ³1. For example, 5 x −1 is a binomial of degree 5.
(iv) The given statement is false, because zero of polynomial can be any real number.
(v) The given statement is false, because a polynomial can have any number of zeroes which depends on the degree of the polynomial.
(vi) The given statement is false. For example, consider the two polynomial −x5 + 3x2 + 4 and x5 + x4 + 2x3 + 3. The degree of each of these polynomial is 5. Their sum is x4 + 2x3 + 3x2 + 7. The degree of this polynomial is not 5.

1. Classify the following polynomial as polynomials in one variable, two variable etc.
(i) x2 + x +1
(ii) y3 −5y
(iii) xy + yz + zx
(iv) x2 − 2xy + y2 +1
Sol. (i) x2 + x +1is a polynomial in one variable.
(ii) y3 −5y is a polynomial in one variable.
(iii) xy + yz + zx is a polynomial in three variable.
(iv) x2 − 2xy + y2 +1 is a polynomial in three variable.

2. Determine the degree of each of the following polynomials:
(i) 2x – 1
(ii) – 10
(iii) x3 −9x + 3x5
(iv) y3 (1− y4)
Sol. (i) Since the highest power of x is 1, the degree of the polynomial 2x – 1 is 1.
(ii) – 10 is a non-zero constant. A non-zero constant term is always regarded as having degree 0.
(iii) Since the highest power of x is 5, the degree of the polynomial x3 −9x + 3x5 is 5.
(iv) y3 (1− y4) = y3 − y7 Since the highest power of y is 7, the degree of the polynomial is 7.

3. For the polynomial

(i) the degree of the polynomial
(ii) the coefficient of 3x .
(iii) the coefficient of 6x .
(iv) the constant term.
Sol. (i) We know that highest power of variable in a polynomial is the degree of the polynomial.
In the given polynomial, the term with highest of x is 6 −x , and the exponent of x in this term in 6.
(ii) The coefficient of 3 x is 1/5
(iii) The coefficient of 6 x is – 1.
(iv) The constant term is 1/5

4. Write the coefficient of 2 x in each of the following:
(i) π/6 x + x2 – 1
(ii) 3x − 5
(iii) (x −1) (3x − 4)
(iv) (2x −5) (2x2 −3x +1)
Sol. (i) The coefficient of 2 x in the given polynomial is 1.
(ii) The given polynomial can be written as 0. x2 + 3x −5. So, the coefficient of x2 in the given polynomial is 0.
(iii) The given polynomial can be written as:
(x −1) (3x − 4) = 3x2 − 4x −3x + 4
= 3x2 − 7x + 4
So, coefficient of 2 x in the given polynomial is 3.
(iv) The given polynomial can be written as:
(2x −5) (2x2 −3x +1) = 4x3 − 6x2 + 2x −10x2 +15x −5
= 4x3 −16x2 +17x −5
So, the coefficient of x2 in the given polynomial is – 16.

5. Classify the following as a constant, linear quadratic and cubic polynomials:
(i) 2 − x2 + x3
(ii) 3x3
(iii) 5t − √7
(iv) 4 − 5y2
(v) 3
(vi) 2 + x
(vii) y3 − y
(viii) 1 + x + x3
(ix) t2
(x) √2x −1
Sol. We know that
(a) a polynomial in which exponent of the variable is zero, is called a constant term.
Here, (v) 3 is a constant polynomial because 3 = 3x0 , exponent of the variable x is 0.
(b) a polynomial of degree 1 is called a linear polynomial.
5t − √7, 2 + x and √2x +1 are linear polynomial.
(c) A polynomial of degree 2 is called a quadratic polynomial.
4 − 5y2 ,1 + x + x2 and t2 are quadratic polynomials.
(d) A polynomial of degree 3 is called a cubic polynomial.
2 − x + x3 ,3x3 and y3 − y are cubic polynomials.

6. Give an example of a polynomial, which is:
(i) monomial of degree 1.
(ii) binomial of degree 20.
(iii) trinomial of degree 2.
Sol. We know that a polynomial having only one term is called a monomial, a polynomial having only two terms is called binomial, a polynomial having only three terms is called a trinomial.
(i) 3x is monomial of degree 1.
(ii) x20 − 7 is a binomial of degree 20.
(iii) 5x2 + 3x −1 is a trinomial of degree 2.

7. Find the value of the polynomial 3x3 − 4x2 + 7x + 5, when x = 3 and also when x = – 3.
Sol. Let p(x) = 3x2 − 4x2 + 7x −5
∴ p(3) = 3(3)3 − 4(3)2 + 7(3) −5
= 3(27) – 4(9) + 21 – 5
= 81 – 36 + 21 – 5
= 61
Now, p(−3) = 3(−3)3 − 4(−3)2 + 7(−3) −5
= 3(−27) − 4(9) − 21− 5
= – 81 – 36 – 21 – 5
= – 143

8. If p(x) = x2 − 4x + 3, evaluate p(2) – p(-1) – p (1/2)
Sol. We have p(x) = x2 − 4x + 3

9. Find p (0), p(1), p(- 2) for the following polynomials:
(i) p(x) =10x − 4x2 −3
(ii) p(y) = (y + 2)(y − 2)
Sol.
(i) We have p(x) =10x − 4x2 −3
∴ p(0) =10(0) − 4(0)2 −3
= 0 − 0 − 3 = −3
And, p(1) =10(1) − 4(1)2 − 3
= 10 − 4 − 3 = 10 − 7 = 3
And, P(−2) =10(−2) − 4(−2)2 −3
= −20 − 4(4) −3 = −20 −16 −3 = −39
(ii) We have p(y) = (y + 2)(y − 2) = y2 − 4
∴ p(0) = (0)2 − 4
= 0 − 4 = −4
And, p(1) = (1)2 − 4
= 1− 4 = −3
And, p(−2) = (−2)2 − 4
= 4 − 4 = 0

10. Verify whether the following are true or false.
(i) – 3 is a zero of x – 3.
(ii) -(1/3) is a zero of 3x +1.
(iii) -(4/5) is a zero of 4 – 5y.
(iv) 0 and 2 are the zeroes of t2 − 2t.
(v) −3 is a zero of y2 + y − 6.
Sol.
A zero of a polynomial p(x) is a number c such that p(c) = 0
(i) Let p(x) = x −3
∴ p(−3) = −3−3 = −6 ≠ 0
Hence, – 3 is not a zero of x – 3.
(ii) Let p(x) = 3x +1

(iv) Let p(t) = t2 − 2t
∴ p(0) = (0)2 − 2(0) = 0
And p(2) = (2)2 − 2(2) = 4 − 4 = 0
Hence, 0 and 2 are zeroes of the polynomial 2 p(t) = t − 2t.
(v) Let p(y) = y2 + y − 6
∴ p(−3) = (−3)2 + (−3) − 6 = 9 −3− 6 = 0
Hence, – 3 is a zero of the polynomial y2 + y − 6.

11. Find the zeroes of the polynomial in each of the following:
(i) p(x) = x − 4
(ii) g(x) = 3− 6x
(iii) q(x) = 2x − 7
(iv) h(y) = 2y
Sol. (i) Solving the equation p(x) = 0, we get x − 4 = 0, which give us x = 4
So, 4 is a zero of the polynomial x − 4.
(ii) Solving the equation g(x) = 0, we get 3− 6x = 0, which gives us x = 1/2
SO,1/2is a zero of the polynomial 3− 6x.
(iii) Solving the equation q(x) = 0, we get 2x − 7 = 0, which gives us x = 7/2
So, 7/2 is a zero of the polynomial 2x − 7.
(iv) Solving the equation h(y) = 0, we get 2y = 0, which gives us y = 0
So, 0 is a zero of the polynomial 2y.

12. Find the zeroes of the polynomial (x − 2)2 − (x + 2)2 . Sol. Let p(x) = (x − 2)2 − (x + 2)2 As fining a zero of p(x), is same as solving the equation p(x) = 0
So, p(x) = 0 ⇒ (x − 2)2 − (x + 2)2 = 0
⇒ (x − 2 + x + 2) (x − 2 − x − 2) = 0
⇒ 2x(−4) = 0 ⇒ −8x = 0 ⇒ x = 0
Hence, x = 0 is the only one zero of p(x).

13. By acute division, find the quotient and the remainder when the first polynomial is divided by the second polynomial: x4 +1; x + 1.
Sol. By acute division, we have

14. By remainder Theorem find the remainder, when p(x) is divided by g(x), where
(i) p(x) = x3 − x22 − 4x −1, g(x) = x +1
(ii) p(x) = x3 −3x2 + 4x + 50, g(x) = x −3
(iii) p(x) = 4×3 −12x2 +14x −3, g(x) = 2x −1
(iv) p (x) = x3 − 6x2 + 2x − 4,g (x) = 1− (3/2)x
Sol. (i) We have g(x) = x + 1
⇒ x +1 = 0
⇒ x = − 0
Remainder = p ( -1)
= (−1)3 − 2(−1)2 − 4(−1) = −1− 2 + 4 −1
= 0
(ii) We have g(x) = x −3
⇒ x − 3 = 0
⇒ x = 3
Remainder = p(3)
= (3)3 −3(3)2 + 4(3) + 50 = 27 − 27 +12 + 50
= 62
(iii) We have g(x) = 2x −1
⇒ 2x −1 = 0
⇒ 2x − 1⇒ x = 1/2
Remainder = p (1/2)

15. Check whether p(x) is a multiple of g(x) or not:
(i) p(x) = x3 − 5x2 + 4x −3, g(x) = x − 2
(ii) p(x) = 2x3 −11x2 − 4x + 5, g(x) = 2x +1
Sol. (i) p(x) will be a multiple g(x) if g(x) divides p(x).
Now, g(x) = x − 2 gives x = 2
Remainder = p(2) = (2)3 − 5(2)2 + 4(2) − 3
= 8 − 5(4) + 8 – 3 = 8 − 20 + 8 − 3
= −7
Since remainder ≠ 0, So p(x) is not a multiple of g(x).
(ii) p(x) will be a multiple of g(x) if g(x) divides p(x).
Now, g(x) = 2x + 1 give x = -(1/2)

16. Show that:
(i) x +3 is a factor of 69 +11x − x2 + x3
(ii) 2x − 3 is a factor of x + 2x3 − 9x2 +12.
Sol.
(i) Let p(x) = 69 + 11x − x2 + x3 , g(x) = x + 3.
g(x) = x + 3 = 0 gives x = −3
g(x) will be a factor of p(x) if p(−3) = 0 (Factor theorem)
Now, p(−3) = 69 +11(−3) − (−3)2 + (−3)3
= 69 − 33 − 9 − 27
= 0
Since, p(−3) = 0, So g(x) is a factor of p(x).
(ii) Let p(x) = x + 2x3 − 9x2 +12 and g(x) = 2x −3
g(x) = 2x −3 = 0 gives x = 3/2

17. Determine which of the following polynomials has x– 2 a factor:
(i) 3x2 + 6x − 24
(ii) 4x2 + x − 2
Sol.
We know that if (x − a) is a factor of p(x), then p(a) = 0.
(i) Let P(x) = 3x2 + 6x − 24
If x – 2 is a factor of p(x) = 3x2 + 6x − 24, then p(2) should be equal to 0.
Now, p(2) = 3(2)2 + 6(2) − 24
= 3(4) + 6(2) − 24
= 12 + 12 – 24
= 0
∴ By factor theorem, (x − 2) is factor of 3x2 + 6x − 24.
(ii) Let 2 p(x) = 4x + x − 2.
If x – 2 is a factor of p(x) = 4x2 + x − 2, then, p(2) should be equal to 0.
Now, p(2) = 4(2)2 + 2 − 2
= 4(4) + 2 − 2
= 16 + 2 – 2
= 16 ≠ 0
∴ x − 2 is not a factor of 4x2 + x − 2.

18. Show that p – 1 is a factor of p10 −1 and also of p11 −1.
Sol. If p −1 is a factor of p10 −1, then (1)10 −1 should be equal to zero.
Now, (1)10 − 1 = 1 − 1 = 0
Therefore, p −1 is a factor of p10 −1.
Again, if p – 1 is a factor of p11 −1, then (1)11 −1 should be equal to zero.
Now, (1)11 − 1 = 1 − 1 = 0
Therefore, p – 1 is a factor of p11 − 1.
Hence, p – 1 is a factor of p10 −1 and also of p11 −1.

19. For what value of m is x3 − 2mx2 + 16 divisible by x + 2?
Sol. If x3 − 2mx2 + 16 is divisible by x + 2, then x + 2 is a factor of x3 − 2mx2 +16.
Now, let p(x) = x3 − 2mx2 + 16.
As x + 2 = x − (−2) is a factor of x3 − 2mx2 + 16.
So p(−2) = 0
Now, p(−2) = (−2) − 2m(−2) +16
= − 8 − 8m + 16 = 8 − 8m
Now, p(−2) = 0
⇒ 8 − 8m = 0
⇒ m = 8 ÷ 8
⇒ m = 1
Hence, for m + 1, x + 2 is a factor of x3 − 2mx2 + 16, so x2 − 2mx2 + 16 is completely
divisible by x +2.

20. If x + 2a is a factor of x3 − 4a2 x3 + 2x + 2a + 3, find a.
Sol. Let p(x) = x5 − 4a2 x3 + 2x + 2a + 3
If x − (−2a) is a factor of p(x), then p(−2a) = 0
∴ p(−2a) = (−2a)5 − 4a2 (−2a)3 + (−2a) + 2a + 3
= −32a5 + 32a5 − 4a + 2a + 3
= −2a + 3
Now, p(−2a) = 0
⇒ −2a + 3 = 0
⇒ a = 3/2

21. Find the value of m so that 2x – 1 be a factor of 8x4 + 4x3 −16x + 10x + m.
Sol. Let p(x) = 8x4 + 4x2 −16x2 + 10x + m.
As (2x −1) is a factor of p(x)

22. If x + 1 is a factor of ax3 + x2 − 2x + 4a −9, find the value of a.
Sol. Let p(x) = ax3 + x2 − 2x + 4a −9.
As (x + 1) is a factor of p(x)
∴ p(−1) = 0 [By factor theorem]
⇒ a(−1)3 + (−1)2 − 2(−1) + 4a −9 = 0
⇒ a(−1) +1+ 2 + 4a −9 = 0
⇒ −a + 4a − 6 = 0
⇒ 3a − 6 = 0 ⇒ 3a = 6 ⇒ a = 2

23. Factorise:
(i) x2 + 9x +18
(ii) 6x2 + 7x −3
(iii) 2x2 − 7x −15
(iv) 84 − 2r − 2r2
Sol. (i) In order to factorise 2 x + 9x +18, we have to find two numbers p and q such that p + q = 9 and pq = 18.
Clearly, 6 + 3 = 9 and 6×3 = 18.
So, we write the middle term 9x as 6x + 3.
∴ x2 + 9x +18 = x2 + 6x + 3x +18
= x(x + 6) + 3(x + 6)
= (x + 6)(x + 3)
(ii) In order to factorise 6x2 + 7x −3,we have to find two numbers p and q such that p + q = 7 and pq = −18.
Clearly, 9 + (−2) = 7 and 9×(−2) = −18.
So, we write the middle term 7x as 9x + (−2x), i.e., 9x − 2x.
∴ 6x2 + 7x −3 = 6x2 + 9x − 2x −3
= 3x(2x + 3) −1(2x + 3)
= (2x + 3)(3x −1)
(iii) In order to factorise 3x2 − 7x −15,we have to find two numbers p and q such that p + q = −7 and pq = −30.
Clearly, (−10) + 3 = −7 and (−10)×3 = −30.
So, we write the middle term −7x as (−10x) +3x.
∴ 2x2 − 7x − 15 = 2x2 −10x + 3x −15
= 2x(x −5) + 3(x −5)
(x −5)(2x + 3)
(iv) In order to factorise 84 − 2r − 2r2 , we have to find two numbers p and 1 such that p + q = −2 and pq = – 168.
84 − 2r − 2r2 = − 2r2 − 2r + 84
= −2r2 −14r +12r +84
= −2r(r + 7) +12(r + 7)
= (r + 7)(−2r +12)
= −2(r + 7)(r − 6) = −2(r − 6)(r + 7)

24. Factorise:
(i) 2×3 −3x2 −17x + 30
(ii) x3 − 6x2 +11x − 6
(iii) x3 + x2 − 4x + 4
(iv) 3×3 − x2 −3x +1
Sol. (i) Let f (x) = 2x3 − 3x2 − 17x + 30 be the given polynomial. The factors of the constant term
+30 are ±1,±2,±3,±5,±6,±10,±15,±30. The factor of coefficient of x3 is 2. Hence, possible rational roots of f(x) are:
±1 ±3 ±5 ±15 ± 1/2 ± 3/2 ± 5/2 ± 15/2
We have f (2) = 2(2)3 −3(2)2 −17(2) + 30
= 2(8) −3(4) −17(2) + 30
= 16 −12 − 34 + 30 = 0
And f (−3) = 2(−3)3 −3(−3)2 −17(−3) + 30
= 2(−27) −3(9) −17(−3) + 30
= −54 − 27 + 51+ 30 = 0
So, (x − 2) and (x +3) are factors of f (x).
⇒ x2 + x − 6 is a factor of f (x).
Let us now divide f (x) = 2x3 − 3x2 −17x +30 by 2 x + x − 6 to get the other factors of f (x). Factors of f (x).
By long division, we have

2x3 − 3x2 − 17x + 30 = (x2 + x − 6)(2x − 5)
⇒ 2x3 − 3x2 − 17x + 30 = (x − 2)(x + 3)(2x − 5)
Hence, 2x3 − 3x2 − 17x + 30 = (x − 2)(x + 3)(2x −5)
(ii) Let f (x) = x3 − 6x2 + 11x + 6 be the given polynomial. The factors of the constant term
– 6 are ±1,±2, ±3 and ±6.
We have, f (1) = (1)3 − 6(1)2 +11(1) − 6 =1− 6 +11− 6 = 0
And, f (2) = (2)3 − 6(2)2 +11(2) − 6 = 8− 24 + 22 = 6 = 0
So, (x −1) and (x − 2) are factor of f (x).
⇒ (x −1)(x − 2) is also factor of f (x).
⇒ x3 −3x + 2 is a factor of f (x).
Let us now divide f (x) = x3 − 6x2 +11x − 6 by x2 − 3x + 2 to get the other factors of f (x).
By long division, we have

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