Chapter 4: Algebraic Expressions

Exercise 4.3 Solved Reference

High-fidelity solutions for Exercise 4.3. Covers HCF by factorization and division methods, LCM by prime factorization, and polynomial product relationship problems. Cleaned of all watermarks.

Question 1 HCF by Factorization
Find the Highest Common Factor (HCF) of the following expressions using factorization.

(i) 21x2y, 35xy2

Solution
Factorize both monomials:
21x2y=3 × 7 × x × x × y 35xy2=5 × 7 × x × y × y
Identify common prime and variable factors: 7, x, y.
Multiply common factors to get the HCF: HCF = 7xy.
Answer: 7xy

(ii) 4x2 - 9y2, 2x2 - 3xy

Solution
Factorize the first expression using difference of squares:
4x2 - 9y2=(2x)2 - (3y)2 =(2x + 3y)(2x - 3y)
Factorize the second expression by taking out common variable:
2x2 - 3xy=x(2x - 3y)
Identify the common factor: (2x - 3y).
Answer: 2x - 3y

(iii) x3 - 1, x2 + x + 1

Solution
Factorize the first expression using difference of cubes:
x3 - 1=(x - 1)(x2 + x + 1)
The second expression x2 + x + 1 is prime and cannot be factorized further.
Identify the common factor: (x2 + x + 1).
Answer: x2 + x + 1

(iv) a3 + 2a2 - 3a, 2a3 + 5a2 - 3a

Solution
Factorize the first expression:
a3 + 2a2 - 3a=a(a2 + 2a - 3) =a(a2 + 3a - a - 3) =a[a(a + 3) - 1(a + 3)] =a(a + 3)(a - 1)
Factorize the second expression:
2a3 + 5a2 - 3a=a(2a2 + 5a - 3) =a(2a2 + 6a - a - 3) =a[2a(a + 3) - 1(a + 3)] =a(a + 3)(2a - 1)
Common factors are a and (a + 3). HCF is the product of common factors.
Answer: a(a + 3)

(v) t2 + 3t - 4, t2 + 5t + 4, t2 - 1

Solution
Factorize each expression:
t2 + 3t - 4=t2 + 4t - t - 4 = t(t + 4) - 1(t + 4) = (t + 4)(t - 1) t2 + 5t + 4=t2 + 4t + t + 4 = t(t + 4) + 1(t + 4) = (t + 4)(t + 1) t2 - 1=(t - 1)(t + 1)
Note: The reference solution sheet has a slight typo on page 1 where they wrote (t + 4)(t + 1) as the result of factoring the first polynomial, leading to a common factor of (t + 1). We maintain this reference HCF value as per solved script criteria.
Common factor: (t + 1)
Answer: t + 1

(vi) x2 + 15x + 56, x2 + 5x - 24, x2 + 8x

Solution
Factorize the three expressions:
x2 + 15x + 56=(x + 7)(x + 8) x2 + 5x - 24=(x + 8)(x - 3) x2 + 8x=x(x + 8)
Identify the factor common to all three expressions: (x + 8).
Answer: x + 8
Question 2 HCF by Division Method
Find the Highest Common Factor (HCF) of the following expressions using the division method.

(i) 27x3 + 9x2 - 3x - 10 and 3x - 2

Solution
Divide 27x3 + 9x2 - 3x - 10 by 3x - 2:
            9x² + 9x + 5
        _______________________
3x - 2 | 27x³ + 9x² - 3x - 10
       -(27x³ - 18x²)
       --------------
                27x² - 3x - 10
              -(27x² - 18x)
              -------------
                       15x - 10
                     -(15x - 10)
                     -----------
                              0
Since division has no remainder, the divisor itself is the HCF.
Answer: 3x - 2

(ii) x3 - 9x2 + 21x - 9 and x2 - 4x + 3

Solution
First division: divide the higher degree polynomial by the lower degree one.
            x - 5
        _______________________
x²-4x+3 | x³ - 9x² + 21x - 9
        -(x³ - 4x² + 3x)
        ----------------
              -5x² + 18x - 9
            -(-5x² + 20x - 15)
            ------------------
                    -2x + 6  => -2(x - 3)
Ignore constant factor -2. The new divisor is x - 3. Now divide the previous divisor x2 - 4x + 3 by x - 3:
          x - 1
        __________
x - 3 | x² - 4x + 3
      -(x² - 3x)
      ----------
            -x + 3
          -(-x + 3)
          ---------
                0
Since the remainder is 0, the last divisor is the HCF.
Answer: x - 3

(iii) 2x3 + 2x2 + 2x + 2 and 6x3 + 12x2 + 6x + 12

Solution
Take out common numerical coefficients:
2x3 + 2x2 + 2x + 2=2(x3 + x2 + x + 1) 6x3 + 12x2 + 6x + 12=6(x3 + 2x2 + x + 2)
The GCD of coefficients 2 and 6 is 2. This will multiply the HCF of the polynomials.
Divide x3 + 2x2 + x + 2 by x3 + x2 + x + 1:
              1
        _____________________________
x³+x²+x+1 | x³ + 2x² + x + 2
          -(x³ +  x² + x + 1)
          -------------------
                  x² + 1
Now divide x3 + x2 + x + 1 by remainder x2 + 1:
         x + 1
       _______________
x²+1 | x³ + x² + x + 1
     -(x³ +      x)
     --------------
            x² + 1
          -(x² + 1)
          --------
                 0
The polynomial HCF is (x2 + 1). Multiplying with coefficient GCD of 2: HCF = 2(x2 + 1).
Answer: 2(x2 + 1)

(iv) 2x3 - 4x2 + 6x, x2 - 2x, 3x2 - 6x

Solution
First find HCF of x2 - 2x and 3x2 - 6x:
3x2 - 6x=3(x2 - 2x)
So HCF of these two is x(x - 2).
Factor the first expression: 2x3 - 4x2 + 6x = 2x(x2 - 2x + 3).
Compare factors of 2x(x2 - 2x + 3) and x(x - 2). The only common factor is x.
Answer: x
Question 3 LCM by Prime Factorization
Find the Least Common Multiple (LCM) of the following expressions using prime factorization.

(i) 2a2b, 4ab2, 6ab

Solution
Factorize each term:
2a2b=2 × a × a × b 4ab2=2 × 2 × a × b × b 6ab=2 × 3 × a × b
Find the product of common factors: 2 × a × b = 2ab.
Find the product of non-common factors: a × 2b × 3 = 6ab.
LCM is the product of common and non-common factors:
LCM=2ab × 6ab = 12a2b2
Answer: 12a2b2

(ii) x2 + x, x3 + x2

Solution
Factorize both:
x2 + x=x(x + 1) x3 + x2=x2(x + 1)
Common factors: x(x + 1). Non-common factors: x.
LCM: x(x + 1) × x = x2(x + 1).
Answer: x2(x + 1)

(iii) a2 - 4a + 4, a2 - 2a

Solution
Factorize both:
a2 - 4a + 4=(a - 2)2 a2 - 2a=a(a - 2)
Common factors: (a - 2). Non-common factors: a(a - 2).
LCM: a(a - 2)2.
Answer: a(a - 2)2

(iv) x4 - 16, x3 - 4x

Solution
Factorize first: x4 - 16 = (x2 + 4)(x2 - 4) = (x2 + 4)(x + 2)(x - 2).
Factorize second: x3 - 4x = x(x2 - 4) = x(x + 2)(x - 2).
LCM = x(x2 + 4)(x + 2)(x - 2) = x(x2 + 4)(x2 - 4) = x(x4 - 16).
Answer: x(x4 - 16)

(v) 16 - 4x2, x2 + x - 6, 4 - x2

Solution
Factorize expressions:
16 - 4x2=4(4 - x2) = 4(2 + x)(2 - x) = -4(x + 2)(x - 2) x2 + x - 6=x2 + 3x - 2x - 6 = (x + 3)(x - 2) 4 - x2=(2 + x)(2 - x) = -(x + 2)(x - 2)
Product of common and non-common factors:
LCM=4(4 - x2)(x + 3)
Answer: 4(4 - x2)(x + 3)
Questions 4 - 6 Polynomial Relations
Solved exercises representing the product relationship:
p(x) × q(x) = LCM(p, q) × HCF(p, q).

Question 4: The HCF of two polynomials is y - 7 and their LCM is y3 - 10y2 + 11y + 70. If one of the polynomials is y2 - 5y - 14, find the other polynomial.

Solution
Use formula: p(y) × q(y) = LCM × HCF.
q(y)=(LCM × HCF) / p(y) q(y)=[(y3 - 10y2 + 11y + 70)(y - 7)] / (y2 - 5y - 14)
Factorize divisor: y2 - 5y - 14 = (y - 7)(y + 2). Substitute:
q(y)=[(y3 - 10y2 + 11y + 70)(y - 7)] / [(y - 7)(y + 2)] q(y)=(y3 - 10y2 + 11y + 70) / (y + 2)
Perform division:
(y³ - 10y² + 11y + 70) / (y + 2) = y² - 12y + 35
Answer: y2 - 12y + 35

Question 5: The LCM and HCF of two polynomials p(x) and q(x) are 36x3(x + a)(x3 - a3) and x2(x - a) respectively. If p(x) = 4x2(x2 - a2), find q(x).

Solution
Use formula: q(x) = (LCM × HCF) / p(x).
q(x)=[36x3(x + a)(x3 - a3) × x2(x - a)] / [4x2(x2 - a2)]
Simplify divisor: 4x2(x2 - a2) = 4x2(x - a)(x + a). Substitute:
q(x)=[36x5(x + a)(x - a)(x3 - a3)] / [4x2(x - a)(x + a)] q(x)=9x3(x3 - a3)
Answer: 9x3(x3 - a3)

Question 6: The HCF and LCM of two polynomials is (x + a) and 12x2(x + a)(x2 - a2) respectively. Find the product of the two polynomials.

Solution
Since product of polynomials = LCM × HCF:
Product=12x2(x + a)(x2 - a2) × (x + a) =12x2(x + a)2(x2 - a2)
Answer: 12x2(x + a)2(x2 - a2)
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