FACTORING POLYNOMIALS USING ALGEBRAIC IDENTITIES

Solved Problems

Problem 1 :

Factor :

x² + 6xy + 9y²

Solution : 

x² + 6xy + 9y²  =  x² + 2(x)(3y) + (3y)²

=  (x + 3y)²

=  (x + 3y)(x + 3y)

Problem 2 :

Factor :

4a² - 20ab + 25b²

Solution : 

4a² - 20ab + 25b²  =  2²a² - 20ab + 5²b²

=  (2a)² - 20ab + (5b)²

=  (2a)² - 2(2a)(5b) + (5b)²

=  (2a - 5b)²

=  (2a - 5b)(2a - 5b)

Problem 3 :

Factor :

2x² + 12xy + 18y²

Solution : 

Factor out the GCF.

=  2(x² + 6xy + 9y²)

=  2[x² + 2(x)(3y) + (3y)²]

=  2(x + 3y)²

=  2(x + 3y)(x + 3y) 

Problem 4 :

Factor :

2ab² - 16ab + 32a

Solution : 

Factor out the GCF.

=  2a(b² - 8b + 16)

=  2a[b² - 2(b)(4) + 4²]

=  2a(b - 4)²

Problem 5 :

Factor :

x² - 25y²

Solution : 

x² - 25y²  =  x² - 5²y²

=  x² - (5y²)

=  (x + 5y)(x - 5y)

Problem 6 :

Factor :

9m² - 16n²

Solution : 

9m² - 16y²  =  3²m² - 4²n²

=  (3m)² - (4n²)

=  (3m + 4n)(3m - 4n)

Problem 7 :

Factor :

x⁴ - y⁴

Solution : 

x⁴ - y⁴  =  (x²)² - (y²)²

=  (x² + y²)(x² - y²)

=  (x² + y²)(x + y)(x - y)

Problem 8 :

Factor :

8p⁴ - 18q⁴

Solution : 

Factor out the GCF.

32p⁴ - 162q⁴  =  2(16p⁴ - 81q⁴)

=  2[4²(p²)² - 9²(q²)²]

=  2[(4p²)² - (9q²)²]

=  2(4p² + 9q²)(4p² - 9q²)

=  2(4p² + 9q²)[(2p)² - (3q)²]

=  2(4p² + 9q²)(2p + 3q)(2p - 3q)

Problem 9 :

If a and b are numbers such that (a−4)(b+6) = 0, then what is the smallest possible value of a² + b² ?

Solution :

Given that, 

(a−4)(b+6) = 0

Equating each factor to 0, we get

a - 4 = 0 and b + 6 = 0

a = 4 and b = -6

Applying the values of a and b solved above in a² + b²

a² + b² = 4² + (-6)²

= 16 + 36

= 52

So, the smallest possible value is 52.

Problem 10 :

Let m be an even integer. How many possible values of m satisfy √(m+7) ≤ 3?

(A) One     (B) Two    (C) Three    (D) Four    (E) Five

Solution :

√(m+7) ≤ 3

The value that we choose for m should be lesser than 7, then only we get positive values inside the square root.

When m = -6

√(-6 + 7) ≤ 3

√1 ≤ 3 (True)

When m = -4

√(-4 + 7) ≤ 3

√3 ≤ 3 (True)

When m = -2

√(-2 + 7) ≤ 3

√5 ≤ 3 (True)

When m = 0

√(0 + 7) ≤ 3

√7 ≤ 3 (True)

When m = 2

√(2 + 7) ≤ 3

√9 ≤ 3 (True)

So, there are 5 solutions for m. Answer is option E.

Problem 11 :

 If (x + 1)² = 4 and (x − 1)² = 16, what are the values of x?

(A) −3   (B) −1   (C) 1   (D) 3    (E) 5

Solution :

(x + 1)² = 4 and (x − 1)² = 16

From (x + 1)² = 4

x + 1 = √4

x + 1 = ±2

x + 1= 2 and x + 1 =  -2

x = 1 and x = -3

From (x − 1)² = 16

x - 1 = √16

x - 1 = ±4

x - 1 = 4 and x - 1 = -4

x = 5 and x = -3

The possible values of x are 1, -3 and 5. So, options A, C and E are correct.

Problem 12 :

If x + y + z = 0, and x² + y² + z² = 16, then find the value of xy + yz + xz.

Solution :

x + y + z = 0 ------(1)

x² + y² + z² = 16 --------(2)

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

0² = 16 + 2(xy + yz + zx)

2(xy + yz + zx) = -16

xy + yz + zx = -16/2

xy + yz + zx = -8

So, the value of xy + yz + zx is -8.

Problem 13 :

If x – y = 4 and xy = 21, compute the value of x³ – y³.

Solution :

x – y = 4 -----(1)

xy = 21 -----(2)

x³ – y³ = (x - y)³ - 3xy(x - y)

Applying the values given, we get

= 4³ - 3(21)(4)

= 64 - 252

= -188

So, the value of x³ – y³ is -188.

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