Factor the quadratic expression $7 x^{2} - 45 x - 28$ completely.

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Solution

Factor the quadratic polynomial $7 x^{2} - 45 x - 28$ :
Find the product of the quadratic term $7 x^{2}$ and the constant term $- 28$ , $7 x^{2} \times (- 28) = - 196 x^{2}$ . Through a hit and trial process, we have to find two factors of $- 196 x^{2}$ that add up to give the linear term $- 45 x$ . The two factors that satisfy these conditions are $- 49 x$ and $4 x$ , because their sum is equal to the linear term $- 45 x$ and their product is equal to $- 196 x^{2}$ .
Rewrite $- 45 x$ in the expression $7 x^{2} - 45 x - 28$ as $- 49 x + 4 x$ .
Then factorize by grouping the terms:
$\begin{array}{rcl} 7 x^{2} - 45 x - 28 & = & 7 x^{2} - 49 x + 4 x - 28 \\ = & (7 x^{2} - 49 x) + (4 x - 28) \\ = & 7 x (x - 7) + 4 (x - 7) \\ = & (x - 7) (7 x + 4) \end{array}$
Thus, $7 x^{2} - 45 x - 28$ is completely factorized to obtain $(x - 7) (7 x + 4)$ .

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Factor the quadratic expression 7x2-45x-28 completely.

Factor the quadratic expression $7 x^{2} - 45 x - 28$ completely.