Polynomials Algebraic Identities Questions With Solutions
Chapter – Polynomials
Topic – Algebraic Identities
Question 1. Expand: (x + 5)².
Solution:
Using the identity:
(a + b)² = a² + 2ab + b²
Here, a = x and b = 5.
Therefore,
(x + 5)² = x² + 2(x)(5) + 5²
= x² + 10x + 25
Answer: x² + 10x + 25
Question 2. Expand: (3x − 4)².
Solution:
Using the identity:
(a − b)² = a² − 2ab + b²
Here, a = 3x and b = 4.
Therefore,
(3x − 4)² = (3x)² − 2(3x)(4) + 4²
= 9x² − 24x + 16
Answer: 9x² − 24x + 16
Question 3. Simplify: (2x + 3)(2x − 3).
Solution:
Using the identity:
(a + b)(a − b) = a² − b²
Here, a = 2x and b = 3.
Therefore,
(2x + 3)(2x − 3)
= (2x)² − 3²
= 4x² − 9
Answer: 4x² − 9
Question 4. Expand: (x + 2)(x + 7).
Solution:
Using the identity:
(x + a)(x + b) = x² + (a + b)x + ab
Here, a = 2 and b = 7.
Therefore,
(x + 2)(x + 7)
= x² + (2 + 7)x + (2 × 7)
= x² + 9x + 14
Answer: x² + 9x + 14
Question 5. Expand: (x + 3)³.
Solution:
Using the identity:
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here, a = x and b = 3.
Therefore,
(x + 3)³
= x³ + 3(x²)(3) + 3(x)(3²) + 3³
= x³ + 9x² + 27x + 27
Answer: x³ + 9x² + 27x + 27
Question 6. Expand: (2x − 1)³.
Solution:
Using the identity:
(a − b)³ = a³ − 3a²b + 3ab² − b³
Here, a = 2x and b = 1.
Therefore,
(2x − 1)³
= (2x)³ − 3(2x)²(1) + 3(2x)(1²) − 1³
= 8x³ − 12x² + 6x − 1
Answer: 8x³ − 12x² + 6x − 1
Question 7. Factorise: x³ + 27 using an algebraic identity.
Solution:
Given:
x³ + 27
We know that:
27 = 3³
Using the identity:
a³ + b³ = (a + b)(a² − ab + b²)
Therefore,
x³ + 27 = x³ + 3³
= (x + 3)(x² − 3x + 9)
Answer: (x + 3)(x² − 3x + 9)
Question 8. Factorise: 8x³ − 125 using an algebraic identity.
Solution:
Given:
8x³ − 125
We can write:
8x³ = (2x)³
125 = 5³
Using the identity:
a³ − b³ = (a − b)(a² + ab + b²)
Therefore,
8x³ − 125
= (2x)³ − 5³
= (2x − 5)(4x² + 10x + 25)
Answer: (2x − 5)(4x² + 10x + 25)
Question 9. Expand: (x + y + z)².
Solution:
Using the identity:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Here, a = x, b = y and c = z.
Therefore,
(x + y + z)²
= x² + y² + z² + 2xy + 2yz + 2zx
Answer: x² + y² + z² + 2xy + 2yz + 2zx
Question 10. If x + y = 7 and xy = 10, find the value of x² + y².
Solution:
Given:
x + y = 7
xy = 10
Using the identity:
(x + y)² = x² + 2xy + y²
Rearranging:
x² + y² = (x + y)² − 2xy
Substitute the given values:
x² + y² = 7² − 2(10)
= 49 − 20
= 29
Answer: 29