Asked by pp
Factor completely.
5, x, squared, minus, 36, x, plus, 7
5x
2
−36x+7
5, x, squared, minus, 36, x, plus, 7
5x
2
−36x+7
Answers
Answered by
GPT-4o mini
To factor the quadratic expression \(5x^2 - 36x + 7\), we can use the method known as "factoring by grouping" or the quadratic formula. However, in this case, since the leading coefficient is not 1, we'll follow the more structured approach.
Let's start by looking for two numbers that multiply to \(5 \cdot 7 = 35\) (the product of the leading coefficient and the constant term) and add to \(-36\) (the coefficient of \(x\)).
The two numbers that fit this description are \(-35\) and \(-1\), since:
\[
-35 \times -1 = 35 \quad \text{and} \quad -35 + -1 = -36.
\]
Now, we can rewrite the middle term of the quadratic:
\[
5x^2 - 35x - 1x + 7.
\]
Next, we can group the terms:
\[
(5x^2 - 35x) + (-1x + 7).
\]
Now, factor out the common factors in each group:
\[
5x(x - 7) - 1(x - 7).
\]
Notice that \((x - 7)\) is a common factor:
\[
(5x - 1)(x - 7).
\]
Thus, the completely factored form of the expression \(5x^2 - 36x + 7\) is:
\[
\boxed{(5x - 1)(x - 7)}.
\]
Let's start by looking for two numbers that multiply to \(5 \cdot 7 = 35\) (the product of the leading coefficient and the constant term) and add to \(-36\) (the coefficient of \(x\)).
The two numbers that fit this description are \(-35\) and \(-1\), since:
\[
-35 \times -1 = 35 \quad \text{and} \quad -35 + -1 = -36.
\]
Now, we can rewrite the middle term of the quadratic:
\[
5x^2 - 35x - 1x + 7.
\]
Next, we can group the terms:
\[
(5x^2 - 35x) + (-1x + 7).
\]
Now, factor out the common factors in each group:
\[
5x(x - 7) - 1(x - 7).
\]
Notice that \((x - 7)\) is a common factor:
\[
(5x - 1)(x - 7).
\]
Thus, the completely factored form of the expression \(5x^2 - 36x + 7\) is:
\[
\boxed{(5x - 1)(x - 7)}.
\]