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Answer :
We are given the quadratic expression
[tex]$$
2c^2 - 17c - 9.
$$[/tex]
To factor this expression, we start by finding two numbers whose product equals the product of the coefficient of [tex]$c^2$[/tex] and the constant term. That is, we need two numbers that multiply to
[tex]$$
2 \times (-9) = -18,
$$[/tex]
and at the same time add to the coefficient of [tex]$c$[/tex], which is [tex]$-17$[/tex]. The numbers that satisfy these requirements are [tex]$-18$[/tex] and [tex]$1$[/tex], since
[tex]$$
-18 \times 1 = -18 \quad \text{and} \quad -18 + 1 = -17.
$$[/tex]
Next, we rewrite the middle term [tex]$-17c$[/tex] as a sum of [tex]$-18c$[/tex] and [tex]$+1c$[/tex]. This gives us
[tex]$$
2c^2 - 18c + c - 9.
$$[/tex]
Now, we use the method of grouping. First, group the terms as follows:
[tex]$$
(2c^2 - 18c) + (c - 9).
$$[/tex]
Factor out the common factors from each group:
1. In the first group, [tex]$2c^2 - 18c$[/tex], the common factor is [tex]$2c$[/tex]:
[tex]$$
2c^2 - 18c = 2c(c - 9).
$$[/tex]
2. In the second group, [tex]$c - 9$[/tex], the common factor is [tex]$1$[/tex]:
[tex]$$
c - 9 = 1(c - 9).
$$[/tex]
Now, since both groups contain the common binomial factor [tex]$(c - 9)$[/tex], we can factor it out:
[tex]$$
2c(c - 9) + 1(c - 9) = (c - 9)(2c + 1).
$$[/tex]
Thus, the factorization of the quadratic is
[tex]$$
2c^2 - 17c - 9 = (c - 9)(2c + 1).
$$[/tex]
So, the correct response is:
A. [tex]$$2c^2-17c-9 = (c-9)(2c+1).$$[/tex]
[tex]$$
2c^2 - 17c - 9.
$$[/tex]
To factor this expression, we start by finding two numbers whose product equals the product of the coefficient of [tex]$c^2$[/tex] and the constant term. That is, we need two numbers that multiply to
[tex]$$
2 \times (-9) = -18,
$$[/tex]
and at the same time add to the coefficient of [tex]$c$[/tex], which is [tex]$-17$[/tex]. The numbers that satisfy these requirements are [tex]$-18$[/tex] and [tex]$1$[/tex], since
[tex]$$
-18 \times 1 = -18 \quad \text{and} \quad -18 + 1 = -17.
$$[/tex]
Next, we rewrite the middle term [tex]$-17c$[/tex] as a sum of [tex]$-18c$[/tex] and [tex]$+1c$[/tex]. This gives us
[tex]$$
2c^2 - 18c + c - 9.
$$[/tex]
Now, we use the method of grouping. First, group the terms as follows:
[tex]$$
(2c^2 - 18c) + (c - 9).
$$[/tex]
Factor out the common factors from each group:
1. In the first group, [tex]$2c^2 - 18c$[/tex], the common factor is [tex]$2c$[/tex]:
[tex]$$
2c^2 - 18c = 2c(c - 9).
$$[/tex]
2. In the second group, [tex]$c - 9$[/tex], the common factor is [tex]$1$[/tex]:
[tex]$$
c - 9 = 1(c - 9).
$$[/tex]
Now, since both groups contain the common binomial factor [tex]$(c - 9)$[/tex], we can factor it out:
[tex]$$
2c(c - 9) + 1(c - 9) = (c - 9)(2c + 1).
$$[/tex]
Thus, the factorization of the quadratic is
[tex]$$
2c^2 - 17c - 9 = (c - 9)(2c + 1).
$$[/tex]
So, the correct response is:
A. [tex]$$2c^2-17c-9 = (c-9)(2c+1).$$[/tex]
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Rewritten by : Barada
