Chapter 4: Algebraic Expressions

Exercise 4.4 Solved Reference

High-fidelity, step-by-step solved reference guide for Exercise 4.4, containing square root extractions of polynomials via factorization and division methods, and polynomial-based optimization word problems.

Question 1 Square Root by Factorization
Find the square root of the following polynomials by factorization.

(i) x2 - 8x + 16

Solution
Write the expression: x2 - 8x + 16
Write as a perfect square:
x2 - 8x + 16=(x)2 - 2(x)(4) + (4)2 =(x - 4)2
Take square root of both sides: √(x2 - 8x + 16) = ±(x - 4).
Answer: ±(x - 4)

(ii) 9x2 + 12x + 4

Solution
Write the expression: 9x2 + 12x + 4
Rewrite as squares: (3x)2 + 2(3x)(2) + (2)2.
Factor using (a + b)2 = a2 + 2ab + b2:
9x2 + 12x + 4=(3x + 2)2
Take the square root: ±(3x + 2).
Answer: ±(3x + 2)

(iii) 36a2 + 84a + 49

Solution
Write expression: 36a2 + 84a + 49
Rewrite components: (6a)2 + 2(6a)(7) + (7)2.
Factor: (6a + 7)2.
Take square root: ±(6a + 7).
Answer: ±(6a + 7)

(iv) 64y2 - 32y + 4

Solution
Write expression: 64y2 - 32y + 4
Factor using identity (a - b)2 = a2 - 2ab + b2:
64y2 - 32y + 4=(8y)2 - 2(8y)(2) + (2)2 =(8y - 2)2
Take square root: ±(8y - 2).
Answer: ±(8y - 2)

(v) 200t2 - 120t + 18

Solution
Write expression and pull out common factor 2:
200t2 - 120t + 18=2(100t2 - 60t + 9)
Factorize perfect square trinomial inside brackets:
100t2 - 60t + 9=(10t)2 - 2(10t)(3) + (3)2 = (10t - 3)2
Combine: 2(10t - 3)2.
Take square root: ±√2(10t - 3).
Answer: ±√2(10t - 3)

(vi) 40x2 + 120x + 90

Solution
Pull out common factor 10:
40x2 + 120x + 90=10(4x2 + 12x + 9)
Factor the inner perfect square trinomial:
4x2 + 12x + 9=(2x)2 + 2(2x)(3) + (3)2 = (2x + 3)2
Take square root: ±√10(2x + 3).
Answer: ±√10(2x + 3)
Question 2 Square Root by Division
Find the square root of the following polynomials by division method.

(i) 4x4 - 28x3 + 37x2 + 42x + 9

Solution
Perform polynomial division:
                2x² - 7x - 3
             ______________________________________
2x²         | 4x⁴ - 28x³ + 37x² + 42x + 9
            |-(4x⁴)
            |------
4x² - 7x    |      -28x³ + 37x²
            |    -(-28x³ + 49x²)
            |    ---------------
4x² - 14x-3 |             -12x² + 42x + 9
            |           -(-12x² + 42x + 9)
            |           ------------------
                                       0
Answer: ±(2x2 - 7x - 3)

(ii) 121x4 - 198x3 - 183x2 + 216x + 144

Solution
Division steps:
                11x² - 9x - 12
             _______________________________________________
11x²        | 121x⁴ - 198x³ - 183x² + 216x + 144
            |-(121x⁴)
            |--------
22x² - 9x   |        -198x³ - 183x²
            |      -(-198x³ + 81x²)
            |      -----------------
22x²-18x-12 |                -264x² + 216x + 144
            |              -(-264x² + 216x + 144)
            |              ---------------------
                                           0
Answer: ±(11x2 - 9x - 12)

(iii) x4 - 10x3y + 27x2y2 - 10xy3 + y4

Solution
Division steps:
                x² - 5xy + y²
             _______________________________________________
x²          | x⁴ - 10x³y + 27x²y² - 10xy³ + y⁴
            |-(x⁴)
            |----
2x² - 5xy   |     -10x³y + 27x²y²
            |   -(-10x³y + 25x²y²)
            |   ------------------
2x²-10xy+y² |              2x²y² - 10xy³ + y⁴
            |            -(2x²y² - 10xy³ + y⁴)
            |            ---------------------
                                        0
Answer: ±(x2 - 5xy + y2)

(iv) 4x4 - 12x3 + 37x2 - 42x + 49

Solution
Division steps:
                2x² - 3x + 7
             ______________________________________
2x²         | 4x⁴ - 12x³ + 37x² - 42x + 49
            |-(4x⁴)
            |------
4x² - 3x    |      -12x³ + 37x²
            |    -(-12x³ + 9x²)
            |    --------------
4x² - 6x + 7|              28x² - 42x + 49
            |            -(28x² - 42x + 49)
            |            ------------------
                                       0
Answer: ±(2x2 - 3x + 7)
Questions 3 - 6 Real-World Applications
Real-world problem applications of quadratic and cubic factorizations.

Question 3: An investor’s return R(x) in rupees after investing x thousand rupees is given by the quadratic expression: R(x) = -x2 + 6x - 8. Factor the expression and find the investment levels that result in zero return.

Solution
Write equation: R(x) = -x2 + 6x - 8.
Factor out negative sign: R(x) = -(x2 - 6x + 8).
Factor the trinomial by middle term split:
R(x)=-(x2 - 4x - 2x + 8) =-[x(x - 4) - 2(x - 4)] =-(x - 4)(x - 2)
To find zero return investment level, set R(x) = 0:
-(x - 4)(x - 2)=0 x = 4orx = 2
Answer: Zero return is achieved at investment levels of 2 thousand rupees and 4 thousand rupees.

Question 4: A company’s profit P(x) in rupees from selling x units of a product is modeled by the cubic expression: P(x) = x3 - 15x2 + 75x - 125. Find the break-even point(s) where the profit is zero.

Solution
Identify cubic expression elements:
P(x)=(x)3 - 3(x)2(5) + 3(x)(5)2 - (5)3
Apply perfect cube identity (a - b)3 = a3 - 3a2b + 3ab2 - b3:
P(x)=(x - 5)3 = (x - 5)(x - 5)(x - 5)
Set profit to zero: (x - 5)3 = 0 \implies x = 5.
Answer: The break-even point is at x = 5 units.

Question 5: The potential energy V(x) in an electric field varies as a cubic function of distance x, given by V(x) = 2x3 - 6x2 + 4x. Determine where the potential energy is zero.

Solution
Write equation and factor out common term 2x:
V(x)=2x(x2 - 3x + 2)
Factor the inner quadratic trinomial:
V(x)=2x(x2 - 2x - x + 2) =2x[x(x - 2) - 1(x - 2)] =2x(x - 2)(x - 1)
Set potential energy to zero: 2x(x - 2)(x - 1) = 0 \implies x = 0, x = 1, x = 2.
Answer: Potential energy is zero at distances x = 0, x = 1, and x = 2.

Question 6: In structural engineering, the deflection y(x) of a beam is given by y(x) = 2x2 - 8x + 6. This equation gives the vertical deflection at any point x along the beam. Find the points of zero deflection.

Solution
Factor out common multiplier 2:
y(x)=2(x2 - 4x + 3)
Factor the trinomial:
y(x)=2(x2 - 3x - x + 3) =2[x(x - 3) - 1(x - 3)] =2(x - 3)(x - 1)
Set deflection to zero: 2(x - 3)(x - 1) = 0 \implies x = 1, x = 3.
Answer: Points of zero deflection along the beam are at x = 1 and x = 3.
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