Chapter 2: Polynomials

Exercise 2.4 - Identity Applications

Exercise 2.4 - Using Identities

❓Question 2: Evaluate without multiplying

(i) 103×107103 \times 107

Step 1: Recognize the pattern 103=100+3103 = 100 + 3 107=100+7107 = 100 + 7

Step 2: Use Identity 4: (x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a+b)x + ab

=(100)2+(3+7)(100)+(3)(7)= (100)^2 + (3 + 7)(100) + (3)(7) =10000+1000+21= 10000 + 1000 + 21 =11021= 11021 ✓

(ii) 95×9695 \times 96

95=100−595 = 100 - 5 96=100−496 = 100 - 4

Using (x−a)(x−b)=x2−(a+b)x+ab(x - a)(x - b) = x^2 - (a+b)x + ab: =10000−900+20=9120= 10000 - 900 + 20 = 9120 ✓

❓Question 7: Evaluate cubes

(i) (99)3(99)^3

Step 1: Write as (100−1)3(100 - 1)^3

Step 2: Use Identity 7: (x−y)3=x3−3x2y+3xy2−y3(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

=(100)3−3(100)2(1)+3(100)(1)2−(1)3= (100)^3 - 3(100)^2(1) + 3(100)(1)^2 - (1)^3 =1000000−30000+300−1= 1000000 - 30000 + 300 - 1 =970299= 970299 ✓

❓Question 14: Use x+y+z=0x + y + z = 0 trick

(i) (−12)3+(7)3+(5)3(-12)^3 + (7)^3 + (5)^3

Step 1: Check if sum = 0 −12+7+5=0-12 + 7 + 5 = 0 ✓

Step 2: Apply special case If x+y+z=0x + y + z = 0, then x3+y3+z3=3xyzx^3 + y^3 + z^3 = 3xyz

=3(−12)(7)(5)= 3(-12)(7)(5) =3(−420)= 3(-420) =−1260= -1260 ✓

(ii) (28)3+(−15)3+(−13)3(28)^3 + (-15)^3 + (-13)^3

Step 1: Check: 28−15−13=028 - 15 - 13 = 0 ✓

Step 2: Apply formula =3(28)(−15)(−13)= 3(28)(-15)(-13) =3×28×195= 3 \times 28 \times 195 =16380= 16380 ✓

Visualizer
Step-by-Step Solution
1
Evaluate 103 × 107
Without direct multiplication
2
103 = 100 + 3, 107 = 100 + 7
Write in (x+a)(x+b) form
3
= (100)² + (3+7)(100) + (3)(7)
Apply Identity 4
4
= 10000 + 1000 + 21
Calculate each part
5
= 11021 ✓
Much faster than multiplying!

Interactive Visualization