Algebra is a branch of mathematics dealing with variables, equations , expressions.Here we are going to solve some of the simple algebra problems including some equations, expressions and algebraic identities.
Solve Simple algebra problems on numbers:
Ex 1: Subtract – 4 from – 10.
Sol :
=– 10 – (– 4)
= – 10 + (additive inverse of – 4)
= – 10 + 4
= – 6 (see addition of two integers)
Ex 2: Find the product of (a) (+ 6) × (– 5) (b) (– 12) × (+ 12) (c) (– 15) × (– 4)
Sol :
(+ 6) × (– 5) = – 30 (positive × negative= negative)
(– 12) × (+ 12) = – 144 (negative × positive = negative)
(– 15) × (– 4)= + 60 (negative × negative = positive)
Ex 3: Find the sum of given algebraic expression 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.
Sol:
Using the associative and distributive property of the real numbers, we obtain (2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1)
= 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1
= 2x4 + 6x3 – (3+6)x2 + (5+4)x + 2
= 2x4 + 6x3 – 9x2 + 9x + 2.
The following scheme is helpful in adding two polynomials 2x4 + 0x3 – 3x2 + 5x + 3 0x4 + 6x3 – 6x2 + 4x – 1 2x4 + 6x3– 9x2 + 9x + 2
Ex 4: Find the subtraction of given simple algebraic expression 2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.
Sol: Using associative and distributive properties, we have (x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1)
= x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1
= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1
= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)
= –x3 + 8x2 – 4x – 5.
The subtraction can also be performed in the following way:
Line (1): x3 + 5x2 – 4x – 6.
Line (2): 2x3 – 3x2 – 1.
Changing the sign of the polynomial in Line (2), we get Line (3): –2x3 + 3x2 + 1.
Adding the polynomial in Line (1) and Line (3), we get –x3 + 8x2 – 4x – 5
Ex 5: Find the product of given simple algebraic expression x3 – 2x2 – 4 and 2x2 + 3x – 1.
Sol:
(x3 – 2x2 – 4) (2x2 + 3x – 1)
= x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–4) (2x2 + 3x – 1)
= (2x5 + 3x4 – x3) + (–4x4 – 6x3 + 2x2) + (–8x2 – 12x + 4)
= 2x5 + 3x4 – x3 – 4x4 – 6x3 + 2x2 – 8x2 – 12x + 4
= 2x5 + (3x4 – 4x4) + (–x3 – 6x3) + (2x2 – 8x2) + (–12x) +4
= 2x5 – x4 – 7x3 – 6x2 – 12x + 4.
Algebra Identities to solve some simple problems :
Algebraic identity for (x + a)(x + b)
By using the distributive properties of numbers,
(x + a )(x + b ) = x(x + b) + a(x + b) = x2 + xb + ax + ab
= x2 + ax + bx + ab= x2 + (a + b)x + ab.
(x + a)(x + b) ≡ x2 + (a + b)x + ab
(x – a)(x + b) ≡ x2 + (b – a)x – ab
(x + a)(x – b) ≡ x2 + (a – b)x – ab
(x – a)(x – b) ≡ x2 – (a + b)x + ab
(a + b)2 ≡ a2 + 2ab + b2
(a – b)2 ≡ a2 – 2ab + b2
(a + b)(a – b) ≡ a2 – b2
(a + b)3 ≡ a3 + 3a2b + 3ab2 + b3
(a − b)3 ≡a3−3a2b+ 3ab2−b3
a3 + b3 ≡ (a + b)3−3ab(a + b)
a3 − b3 ≡ (a −b)3+ 3ab(a−b)
Solve the following simple algebra problems using above identities:
Pro 1: Find the product: (x + 3) (x + 5)
Sol: (x + 3) (x + 5) = x2 + (3 + 5) x + 3 × 5 = x2 + 8x + 15.
Pro 2: Find the product: (p + 9) (p – 2)
Sol: (p + 9) (p – 2) = p2 + (9 – 2) p – 9 × 2
= p2 + 7p – 18.
Solve simple algebra problems on the cubic identity
Algebra problems on cubic identity:
(i) (3x+2y)3 = (3x)3 + 3(3x)2(2y) + 3(3x)(2y)2 + (2y)3
= 27x3 + 3(9x2)(2y) + 3(3x)(4y2) + 8y3
= 27x3 + 54x2y + 36xy2 + 8y3.
(ii) (2x2– 3y)3 = (2x2)3 – 3(2x2)2 (3y) + 3(2x2) (3y)2 – (3y)3
= 8x6 – 3(4x4)(3y) + 3(2x2)(9y2) – 27y3
= 8x6– 36x4y + 54x2y2 – 27y3
Pro 1: If the values of a+b and ab are 4 and 1 respectively, find the value of a3 + b3.
Solution:a3+ b3 = (a+b)3 – 3ab(a+b) = (4)3 – 3(1)(4)
= 64 – 12
= 52.
Pro 2: If a – b = 4 and ab = 2, find a3 – b3.
Solution:a3 – b3 = (a–b)3+3ab(a–b)
= (4)3+3(2)(4) =64+24
=88.
Solve Simple algebra problems on numbers:
Ex 1: Subtract – 4 from – 10.
Sol :
=– 10 – (– 4)
= – 10 + (additive inverse of – 4)
= – 10 + 4
= – 6 (see addition of two integers)
Ex 2: Find the product of (a) (+ 6) × (– 5) (b) (– 12) × (+ 12) (c) (– 15) × (– 4)
Sol :
(+ 6) × (– 5) = – 30 (positive × negative= negative)
(– 12) × (+ 12) = – 144 (negative × positive = negative)
(– 15) × (– 4)= + 60 (negative × negative = positive)
Ex 3: Find the sum of given algebraic expression 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.
Sol:
Using the associative and distributive property of the real numbers, we obtain (2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1)
= 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1
= 2x4 + 6x3 – (3+6)x2 + (5+4)x + 2
= 2x4 + 6x3 – 9x2 + 9x + 2.
The following scheme is helpful in adding two polynomials 2x4 + 0x3 – 3x2 + 5x + 3 0x4 + 6x3 – 6x2 + 4x – 1 2x4 + 6x3– 9x2 + 9x + 2
Ex 4: Find the subtraction of given simple algebraic expression 2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.
Sol: Using associative and distributive properties, we have (x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1)
= x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1
= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1
= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)
= –x3 + 8x2 – 4x – 5.
The subtraction can also be performed in the following way:
Line (1): x3 + 5x2 – 4x – 6.
Line (2): 2x3 – 3x2 – 1.
Changing the sign of the polynomial in Line (2), we get Line (3): –2x3 + 3x2 + 1.
Adding the polynomial in Line (1) and Line (3), we get –x3 + 8x2 – 4x – 5
Ex 5: Find the product of given simple algebraic expression x3 – 2x2 – 4 and 2x2 + 3x – 1.
Sol:
(x3 – 2x2 – 4) (2x2 + 3x – 1)
= x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–4) (2x2 + 3x – 1)
= (2x5 + 3x4 – x3) + (–4x4 – 6x3 + 2x2) + (–8x2 – 12x + 4)
= 2x5 + 3x4 – x3 – 4x4 – 6x3 + 2x2 – 8x2 – 12x + 4
= 2x5 + (3x4 – 4x4) + (–x3 – 6x3) + (2x2 – 8x2) + (–12x) +4
= 2x5 – x4 – 7x3 – 6x2 – 12x + 4.
Algebra Identities to solve some simple problems :
Algebraic identity for (x + a)(x + b)
By using the distributive properties of numbers,
(x + a )(x + b ) = x(x + b) + a(x + b) = x2 + xb + ax + ab
= x2 + ax + bx + ab= x2 + (a + b)x + ab.
(x + a)(x + b) ≡ x2 + (a + b)x + ab
(x – a)(x + b) ≡ x2 + (b – a)x – ab
(x + a)(x – b) ≡ x2 + (a – b)x – ab
(x – a)(x – b) ≡ x2 – (a + b)x + ab
(a + b)2 ≡ a2 + 2ab + b2
(a – b)2 ≡ a2 – 2ab + b2
(a + b)(a – b) ≡ a2 – b2
(a + b)3 ≡ a3 + 3a2b + 3ab2 + b3
(a − b)3 ≡a3−3a2b+ 3ab2−b3
a3 + b3 ≡ (a + b)3−3ab(a + b)
a3 − b3 ≡ (a −b)3+ 3ab(a−b)
Solve the following simple algebra problems using above identities:
Pro 1: Find the product: (x + 3) (x + 5)
Sol: (x + 3) (x + 5) = x2 + (3 + 5) x + 3 × 5 = x2 + 8x + 15.
Pro 2: Find the product: (p + 9) (p – 2)
Sol: (p + 9) (p – 2) = p2 + (9 – 2) p – 9 × 2
= p2 + 7p – 18.
Solve simple algebra problems on the cubic identity
Algebra problems on cubic identity:
(i) (3x+2y)3 = (3x)3 + 3(3x)2(2y) + 3(3x)(2y)2 + (2y)3
= 27x3 + 3(9x2)(2y) + 3(3x)(4y2) + 8y3
= 27x3 + 54x2y + 36xy2 + 8y3.
(ii) (2x2– 3y)3 = (2x2)3 – 3(2x2)2 (3y) + 3(2x2) (3y)2 – (3y)3
= 8x6 – 3(4x4)(3y) + 3(2x2)(9y2) – 27y3
= 8x6– 36x4y + 54x2y2 – 27y3
Pro 1: If the values of a+b and ab are 4 and 1 respectively, find the value of a3 + b3.
Solution:a3+ b3 = (a+b)3 – 3ab(a+b) = (4)3 – 3(1)(4)
= 64 – 12
= 52.
Pro 2: If a – b = 4 and ab = 2, find a3 – b3.
Solution:a3 – b3 = (a–b)3+3ab(a–b)
= (4)3+3(2)(4) =64+24
=88.
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