Pergunta

a^2+4a+4 2. 6x^2+22x+12 3. 8a^2-2a-1 u b^2+4b-5 square 16-49a^2
Solução

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JullyaElite · Tutor por 8 anos
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Let's factor each expression step by step:
1. a^2 + 4a + 4
This is a perfect square trinomial. It can be factored as:
(a + 2)^2
2. 6x^2 + 22x + 12
First, factor out the greatest common factor (GCF), which is 2:
6x^2 + 22x + 12 = 2(3x^2 + 11x + 6)
Next, factor the quadratic inside the parentheses:
3x^2 + 11x + 6
We need to find two numbers that multiply to 3 \times 6 = 18 and add up to 11. These numbers are 9 and 2.
So, we can rewrite the quadratic as:
3x^2 + 9x + 2x + 6
Now, factor by grouping:
= 3x(x + 3) + 2(x + 3)
= (3x + 2)(x + 3)
Thus, the factored form is:
2(3x + 2)(x + 3)
3. 8a^2 - 2a - 1
This quadratic does not factor nicely with integer coefficients. We can use the quadratic formula to find the roots if needed:
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a = 8, b = -2, and c = -1.
4. b^2 + 4b - 5
We need to find two numbers that multiply to -5 and add up to 4. These numbers are 5 and -1.
So, we can rewrite the quadratic as:
b^2 + 5b - b - 5
Now, factor by grouping:
= b(b + 5) - 1(b + 5)
= (b - 1)(b + 5)
Thus, the factored form is:
(b - 1)(b + 5)
5. 16 - 49a^2
This is a difference of squares. It can be factored as:
16 - 49a^2 = (4)^2 - (7a)^2
= (4 - 7a)(4 + 7a)
So, the factored form is:
(4 - 7a)(4 + 7a)
In summary:
1. (a + 2)^2
2. 2(3x + 2)(x + 3)
3. 8a^2 - 2a - 1 (no simple factorization)
4. (b - 1)(b + 5)
5. (4 - 7a)(4 + 7a)
1. a^2 + 4a + 4
This is a perfect square trinomial. It can be factored as:
(a + 2)^2
2. 6x^2 + 22x + 12
First, factor out the greatest common factor (GCF), which is 2:
6x^2 + 22x + 12 = 2(3x^2 + 11x + 6)
Next, factor the quadratic inside the parentheses:
3x^2 + 11x + 6
We need to find two numbers that multiply to 3 \times 6 = 18 and add up to 11. These numbers are 9 and 2.
So, we can rewrite the quadratic as:
3x^2 + 9x + 2x + 6
Now, factor by grouping:
= 3x(x + 3) + 2(x + 3)
= (3x + 2)(x + 3)
Thus, the factored form is:
2(3x + 2)(x + 3)
3. 8a^2 - 2a - 1
This quadratic does not factor nicely with integer coefficients. We can use the quadratic formula to find the roots if needed:
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a = 8, b = -2, and c = -1.
4. b^2 + 4b - 5
We need to find two numbers that multiply to -5 and add up to 4. These numbers are 5 and -1.
So, we can rewrite the quadratic as:
b^2 + 5b - b - 5
Now, factor by grouping:
= b(b + 5) - 1(b + 5)
= (b - 1)(b + 5)
Thus, the factored form is:
(b - 1)(b + 5)
5. 16 - 49a^2
This is a difference of squares. It can be factored as:
16 - 49a^2 = (4)^2 - (7a)^2
= (4 - 7a)(4 + 7a)
So, the factored form is:
(4 - 7a)(4 + 7a)
In summary:
1. (a + 2)^2
2. 2(3x + 2)(x + 3)
3. 8a^2 - 2a - 1 (no simple factorization)
4. (b - 1)(b + 5)
5. (4 - 7a)(4 + 7a)
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