Question

factor the following trinomial. 100x² + 140x + 49 (?x + \square)²

Explanation:

Step1: Identify the form of perfect square trinomial

A perfect square trinomial has the form \(a^{2}+2ab + b^{2}=(a + b)^{2}\). For the trinomial \(100x^{2}+140x + 49\), we first find \(a\) and \(b\) such that \(a^{2}=100x^{2}\) and \(b^{2}=49\).
From \(a^{2}=100x^{2}\), we take the square root: \(a=\sqrt{100x^{2}} = 10x\) (we consider the positive root for the form \((a + b)^{2}\)).
From \(b^{2}=49\), we take the square root: \(b=\sqrt{49}=7\) (positive root as well).

Step2: Verify the middle term

Now, we check if \(2ab\) equals the middle term \(140x\). Substitute \(a = 10x\) and \(b = 7\) into \(2ab\):
\(2ab=2\times(10x)\times7=140x\), which matches the middle term of the trinomial.

Step3: Write the factored form

Since \(100x^{2}+140x + 49=(10x)^{2}+2\times(10x)\times7+7^{2}\), by the perfect square trinomial formula, it factors to \((10x + 7)^{2}\).

Answer:

The first box (for the coefficient of \(x\)) is \(10\) and the second box is \(7\), so the factored form is \((10x + 7)^{2}\).