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Mathematics Polynomials Worksheets for Class 9
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Chapter-wise Class 9 Mathematics Polynomials Worksheets Pdf Download
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.
(d) 4 Biquadratic x4 -1
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
1. Which of the following expression are polynomials? Justify your answer.
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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