Question Factor completely. 5x^2-36x+7 square

To factor the quadratic expression $5x^2 - 36x + 7$ completely, we need to find two binomials whose product gives us the original quadratic expression.

The general form of factoring a quadratic expression $ax^2 + bx + c$ is:
$$(mx + n)(px + q)$$

For our specific quadratic expression $5x^2 - 36x + 7$, we need to find two numbers that multiply to $a \cdot c = 5 \cdot 7 = 35$ and add up to $b = -36$.

Let's denote these two numbers as $m$ and $n$. We need:
$$m \cdot n = 35$$
$$m + n = -36$$

By examining the pairs of factors of 35, we find:
- $1 \cdot 35 = 35$ and $1 + 35 = 36$
- $-1 \cdot -35 = 35$ and $-1 + (-35) = -36$

Thus, the correct pair is $-1$ and $-35$.

Now, we rewrite the middle term $-36x$ using these two numbers:
$$5x^2 - 36x + 7 = 5x^2 - 1x - 35x + 7$$

Next, we group the terms:
$$= (5x^2 - 1x) + (-35x + 7)$$

We factor out the greatest common factor (GCF) from each group:
$$= x(5x - 1) - 7(5x - 1)$$

Now, we factor out the common binomial factor $(5x - 1)$:
$$= (5x - 1)(x - 7)$$

Therefore, the completely factored form of $5x^2 - 36x + 7$ is:
$$\boxed{(5x - 1)(x - 7)}$$