Chapter 4: Algebraic Expressions

Review Exercise 4 Solved Notes

Comprehensive solved reference containing interactive multiple-choice questions, polynomial factorizations, HCF & LCM problems, and real-world loan applications, completely cleaned of all watermarks.

Question 1 Multiple Choice Questions
Select the correct option. Click on an option to check if it's correct.
1. The factorization of 12x + 36 is:
Explanation: Pull out the greatest common divisor of coefficients, which is 12: 12(x + 3).
2. The factors of 4x2 - 12x + 9 are:
Explanation: This is a perfect square trinomial: (2x)² - 2(2x)(3) + 3² = (2x - 3)².
3. The HCF of a2b2 and ab3 is:
Explanation: Take the lowest power of each variable common to both: a1 and b2, giving HCF = ab².
4. The LCM of 16x2, 4x, 30xy is:
Explanation: LCM of coefficients (16, 4, 30) is 240. LCM of variables (x², x, xy) is x²y. Hence, 240x²y.
5. Product of LCM and HCF = __________ of two polynomials.
Explanation: The fundamental relation is p(x) × q(x) = HCF × LCM.
6. The square root of x2 - 6x + 9 is:
Explanation: Since x² - 6x + 9 = (x - 3)², its square root is ±(x - 3).
7. The LCM of (a - b)2 and (a - b)3 is:
Explanation: For LCM, we take the highest power of common terms, which is (a - b)³.
8. Factorization of x3 + 3x2 + 3x + 1 is:
Explanation: Fits perfect cube form: (x)³ + 3(x)²(1) + 3(x)(1)² + (1)³ = (x + 1)³.
9. Cubic polynomial has degree:
Explanation: The degree of a cubic polynomial is 3.
10. One of the factors of x3 - 27 is:
Explanation: x³ - 27 = (x - 3)(x² + 3x + 9). Therefore, (x - 3) is one factor.
Question 2 Factorize Expressions
Factorize the following algebraic expressions.

(i) 4x3 + 18x2 - 12x

Solution
Identify common monomial factor: 2x.
Factor out 2x: 2x(2x2 + 9x - 6).
Answer: 2x(2x2 + 9x - 6)

(ii) x3 + 64y3

Solution
Rewrite as sum of cubes: (x)3 + (4y)3.
Apply identity a3 + b3 = (a + b)(a2 - ab + b2):
x3 + 64y3=(x + 4y)[(x)2 - (x)(4y) + (4y)2] =(x + 4y)(x2 - 4xy + 16y2)
Answer: (x + 4y)(x2 - 4xy + 16y2)

(iii) x3y3 - 8

Solution
Rewrite as difference of cubes: (xy)3 - (2)3.
Apply difference of cubes identity:
x3y3 - 8=(xy - 2)[(xy)2 + 2xy + (2)2] =(xy - 2)(x2y2 + 2xy + 4)
Answer: (xy - 2)(x2y2 + 2xy + 4)

(iv) -x2 - 23x - 60

Solution
Factor out negative sign: -(x2 + 23x + 60).
Split the middle term:
=-(x2 + 20x + 3x + 60) =-[x(x + 20) + 3(x + 20)] =-(x + 20)(x + 3)
Answer: -(x + 20)(x + 3)

(v) 2x2 + 7x + 3

Solution
Factor using middle-term splitting:
2x2 + 7x + 3=2x2 + 6x + x + 3 =2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)
Answer: (2x + 1)(x + 3)

(vi) x4 + 64

Solution
Complete square for (x2)2 + (8)2 by adding and subtracting 2(x2)(8) = 16x2:
x4 + 64=(x2)2 + 82 + 16x2 - 16x2 =(x2 + 8)2 - (4x)2
Apply difference of squares identity:
=(x2 + 8 + 4x)(x2 + 8 - 4x) =(x2 + 4x + 8)(x2 - 4x + 8)
Answer: (x2 + 4x + 8)(x2 - 4x + 8)

(vii) x4 + 2x2 + 9

Solution
Rearrange terms: x4 + 9 + 2x2.
Complete square for (x2)2 + 32 by adding and subtracting 2(x2)(3) = 6x2:
x4 + 9 + 2x2=(x2)2 + 32 + 6x2 - 6x2 + 2x2 =(x2 + 3)2 - 4x2 =(x2 + 3)2 - (2x)2
Factor:
=(x2 + 3 + 2x)(x2 + 3 - 2x) =(x2 + 2x + 3)(x2 - 2x + 3)
Answer: (x2 + 2x + 3)(x2 - 2x + 3)

(viii) (x + 3)(x + 4)(x + 5)(x + 6) - 360

Solution
Group constant sums: 3 + 6 = 9 and 4 + 5 = 9.
=[(x + 3)(x + 6)][(x + 4)(x + 5)] - 360 =[x2 + 9x + 18][x2 + 9x + 20] - 360
Let y = x2 + 9x:
=(y + 18)(y + 20) - 360 =y2 + 38y + 360 - 360 = y2 + 38y =y(y + 38)
Substitute back y = x2 + 9x:
=(x2 + 9x)(x2 + 9x + 38) =x(x + 9)(x2 + 9x + 38)
Answer: x(x + 9)(x2 + 9x + 38)

(ix) (x2 + 6x + 3)(x2 + 6x - 9) + 36

Solution
Let y = x2 + 6x. Substitute:
=(y + 3)(y - 9) + 36 =y2 - 6y - 27 + 36 = y2 - 6y + 9
Factor as perfect square: (y - 3)2.
Substitute back: (x2 + 6x - 3)2.
Answer: (x2 + 6x - 3)2
Question 3 LCM & HCF Problems
Find the LCM and HCF of the following polynomial groups.

(i) 4x3 + 12x2, 8x2 + 16x

Solution
Factorize both:
4x3 + 12x2=4x2(x + 3) = 2 × 2 × x × x × (x + 3) 8x2 + 16x=8x(x + 2) = 2 × 2 × 2 × x × (x + 2)
HCF (common factors): 4x.
LCM: 4x × x(x + 3) × 2(x + 2) = 8x2(x + 3)(x + 2).
Answer: HCF = 4x, LCM = 8x2(x + 3)(x + 2)

(ii) x3 + 3x2 - 4x, x2 - 4x + 3

Solution
Factorize first: x(x2 + 3x - 4) = x(x + 4)(x - 1).
Factorize second: (x - 3)(x - 1).
HCF: (x - 1).
LCM: x(x - 1)(x - 3)(x + 4).
Answer: HCF = x - 1, LCM = x(x - 1)(x - 3)(x + 4)

(iii) x2 + 8x + 16, x2 - 16

Solution
Factorize:
x2 + 8x + 16=(x + 4)2 x2 - 16=(x + 4)(x - 4)
HCF: (x + 4).
LCM: (x + 4)2(x - 4).
Answer: HCF = x + 4, LCM = (x + 4)2(x - 4)

(iv) x3 - 9x, x2 - x - 6

Solution
Factorize:
x3 - 9x=x(x2 - 9) = x(x + 3)(x - 3) x2 - x - 6=(x - 3)(x + 2)
HCF: (x - 3).
LCM: x(x + 2)(x - 3)(x + 3) = x(x + 2)(x2 - 9).
Answer: HCF = x - 3, LCM = x(x + 2)(x2 - 9)
Questions 4 - 5 Square Roots & Loan Cost
Square root computation and real-world loan cost optimization.

Question 4: Find the square root of the expression 16x4 + 8x2 + 1 using factorization and division method.

Solution
Factorization Method:
16x4 + 8x2 + 1=(4x2)2 + 2(4x2)(1) + 12 = (4x2 + 1)2
Taking square root: ±(4x2 + 1).
Division Method:
            4x² + 1
         _______________
4x²      | 16x⁴ + 8x² + 1
         |-(16x⁴)
         |------
8x² + 1  |        8x² + 1
         |      -(8x² + 1)
         |      ----------
                      0
Answer: ±(4x2 + 1)

Question 5: Huria is analyzing the total cost of her loan, modeled by the expression C(x) = x2 - 8x + 15, where x represents the number of years. What is the optimal repayment period for Huria’s loan?

Solution
Factor the cost model expression:
C(x)=x2 - 5x - 3x + 15 =x(x - 5) - 3(x - 5) = (x - 5)(x - 3)
The cost of the loan is minimized (reaches zero) when C(x) = 0.
Set (x - 5)(x - 3) = 0 \implies x = 3 or x = 5.
Answer: The optimal repayment period is 3 years or 5 years.
Interactive Sandbox Suite Comprehensive Math Utilities
Factorize ax² + bx + c by middle term split.
Compute HCF and LCM of two monomials.
Evaluate key identities.