To factor the trinomial \(2x^2 + 5x - 3\), we are looking for two binomials of the form \((ax + b)(cx + d)\) that when multiplied together, result in the original trinomial.
Given the expression \(2x^2 + 5x - 3\):
- The product of \(a\) and \(c\) must equal the coefficient of \(x^2\), which is \(2\).
- The product of \(b\) and \(d\) must equal the constant term, which is \(-3\).
- The sum of the inner and outer products \(ad + bc\) must equal the coefficient of \(x\), which is \(5\).
After analyzing these factors, we can find that \(2x^2 + 5x - 3\) factors to:
\[ (2x - 1)(x + 3) \]
Thus, the expression \(2x^2 + 5x - 3\) is equivalent to the product of the binomials \(2x - 1\) and \(x + 3\).
For the example \(3x^2 + 22x + 7\), similar factorization methods should be applied.
Letβs represent the findings in the table format as per your example:
\[ \begin{array}{|c|c|} \hline 2x^2 + 5x - 3 & 3x^2 + 22x + 7 \ \hline (2x - 1)(x + 3) & \text{(needs to be factored)} \ \hline \end{array} \]
To factor \(3x^2 + 22x + 7\):
- The possible pairs of factors for \(3\) (product of the first term) are \(3\) and \(1\).
- The pairs for \(7\) (constant term) are \(7\) and \(1\).
Trial and error shows the factors of \(3x^2 + 22x + 7\) will lead to: \[ (3x + 1)(x + 7) \]
So we can complete the table with that result:
\[ \begin{array}{|c|c|} \hline 2x^2 + 5x - 3 & 3x^2 + 22x + 7 \ \hline (2x - 1)(x + 3) & (3x + 1)(x + 7) \ \hline \end{array} \]
This is your complete answer!