There are some techniques of polynomial factorization which allow us to write them as a multiplication of two or more polynomials.
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Question 1. Writing the common factor into evidence, factor the polynomials:
a) 15x + 15y
b) x² + 9xy
c) ab – a³b³
d) a²z + abz
Question 2. Factor each of the polynomials:
a) x² – xy – x
b) 24x³ – 8x² – 56x³
c) a.(x + y) – b.(x + y)
d) b.(a – x) – c.(a – x)
Question 3. Using the clustering and common-factor-in-evidence techniques, factor the following polynomials:
a) a² + ab + ax + bx
b) bx² – 2by + 5x² – 10y
c) 2an + n -2am – m
d) ax – bx + cx + ay – by + cy
Question 4. The polynomials below show differences of two squares. Write each of them in factored form.
a) a² – 64
b) (x – 4)² – 16
c) (y + 1)² – 25
d) x² – (x + y)²
Question 5. Factor the following polynomial by writing as a multiplication:
(a – b + 2)² – (a – b – 2)²
Question 6. Check that each of the trinomials below represents a perfect square trinomial, then do the factorization.
a) a² – 10ab + 25b²
b) x² – 8x + 25
c) 9x² – 6x + 1
d) 16a² + 24ab + 9b²
Question 7. Complete the polynomial below so that it is a perfect square trinomial.
x² + 4x
Question 8. Using factoring techniques, find the roots of the equations:
a) x² – 9x = 0
b) x² – 64 = 0
c) y² – y = 0
d) x² – 1 = 0
a) 15x + 15y = 15.(x + y)
b) x² + 9xy = x.(x + 9y)
c) ab – a³b³ = ab.(1 – a²b²)
d) a²z + abz = az.(a + b)
a) x² – xy – x = x.(x – y -1)
b) 24x³ – 8x² – 56x³ = 8x².(3x – 1 – 7x)
c) a.(x + y) – b.(x + y) = (x + y).(a + b)
d) b.(a – x) – c.(a – x) = (a – x).(b – c)
a) a² + ab + ax + bx = a.(a + b) + x (a + b) = (a + b).(a + x)
b) bx² – 2by + 5x² – 10y = bx² + 5x² – 2by – 10y = x².(b + 5) – 2y.(b + 5) = (b + 5).(x² – 2y)
c) 2an + n -2am – m = n.(2a + 1) – m.(2a + 1) = (2a + 1).(n – m)
d) ax – bx + cx + ay – by + cy = x.(a – b + c) + y.(a – b + c) = (a + b + c).(x + y)
a) a² – 64 = (a + 8).(a – 8)
b) (x – 4)² – 16 = ((x – 4) + 4). ((x – 4) – 4) = (x – 4 + 4).(x – 4 – 4) = x.(x – 8)
c) (y + 1)² – 25 = ((y + 1) + 5). ((y + 1) – 5) = (y + 1 + 5).(y + 1 – 5) = (y + 6).(y – 4)
d) x² – (x + y) ² = (x + (x + y)). (x – (x + y)) = (x + x + y).(x – x – y) = (2x + y).(- y) = -y.(2x + y)
(a – b + 2)² – (a – b – 2)² =
((a – b + 2) + (a – b – 2)). ((a – b + 2) – (a – b – 2)) =
(a – b + 2 + a – b – 2). (a – b + 2 – a + b + 2) =
(2a – 2b). (4) =
4.(2a – 2b)
a) a² – 10ab + 25b²
First, we take the square root of the terms we square:
√a² = The
√25b² = 5b
Like 2. The. 5b = 10ab → remaining term of the trinomial. So the polynomial is a perfect square trinomial.
Let's factor: a² – 10ab + 25b² = (a – 5b)²
b) x² – 8x + 25
√x² = x
√25 = 5
2. x. 5 = 10x → does not match the remaining term which is 8x. So the polynomial is not a perfect square trinomial.
c) 9x² – 6x + 1
√9x² = 3x
√1 = 1
2. 3x. 1 = 6x → remaining term of the trinomial. So the polynomial is a perfect square trinomial.
Let's factor: 9x² – 6x + 1 = (3x – 1)²
d) 16a² + 24ab + 9b²
√16a² = 4th
√9b² = 3b
2. 4th. 3b = 24ab → remaining term of the trinomial. So the polynomial is a perfect square trinomial.
Let's factor: 16a² + 24ab + 9b² = (4a + 3b)²
x² + 4x
We must write a perfect square trinomial as follows: x² + 2xy + y² = (x + y)²
So we need to find the value of y. We have:
2xy = 4x
2y = 4
y = 4/2
y = 2
Thus, we must add the term y² = 2² = 4 to the polynomial so that it is a perfect square trinomial: x² + 4x + 4 = (x + 2)².
a) Placing x in evidence:
x.(x – 9) = 0
Then x = 0 or
x – 9 = 0 ⇒ x = 9
Roots: 0 and 9
b) We have a difference between two squares:
x² – 64 = 0
⇒ (x + 8).(x – 8) = 0
That is, x + 8 = 0 or x – 8 = 0.
x + 8 = 0 ⇒ x = -8
x – 8 = 0 ⇒ x = 8
Roots: -8 and 8.
c) Putting y in evidence:
y.(y – 1) = 0
So y = 0 or y – 1 = 0.
y – 1 = 0 ⇒ y = 1
Roots: 0 and 1
d) Remembering that 1 = 1², we have a difference between two squares:
x² – 1 = 0
⇒ (x + 1).(x – 1) = 0
So x + 1 = 0 or x – 1 = 0.
x + 1 = 0 ⇒ x = -1
x – 1 = 0 ⇒ x = 1
Roots: – 1 and 1.
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