Factor the trinomial: \[ 2 x^{2}+15 x+28 \]
Solution
Step 1 ：To factor the trinomial \(2x^2 + 15x + 28\), we need to find two numbers that multiply to \(2*28 = 56\) and add up to \(15\).
Step 2 ：The numbers that satisfy these conditions are \(7\) and \(8\), because \(7*8 = 56\) and \(7+8 = 15\).
Step 3 ：So, we can write the trinomial as: \(2x^2 + 15x + 28 = 2x^2 + 7x + 8x + 28\).
Step 4 ：Now, we can factor by grouping: \(2x^2 + 7x + 8x + 28 = x(2x + 7) + 4(2x + 7)\).
Step 5 ：Since the terms in the parentheses are the same, we can factor them out: \(x(2x + 7) + 4(2x + 7) = (x + 4)(2x + 7)\).
Step 6 ：So, the factored form of the trinomial \(2x^2 + 15x + 28\) is \((x + 4)(2x + 7)\).
Step 7 ：We can check our answer by expanding the factored form: \((x + 4)(2x + 7) = 2x^2 + 7x + 8x + 28 = 2x^2 + 15x + 28\).
Step 8 ：So, our answer is correct. The factored form of the trinomial \(2x^2 + 15x + 28\) is \(\boxed{(x + 4)(2x + 7)}\).
