Question

factor the following trinomial.\\(9x^2 - 42x + 49\\)\\((?x - \square)^2\\)

Explanation:

Step1: Identify the perfect square terms

The given trinomial is \(9x^{2}-42x + 49\). We know that \((ax - b)^{2}=a^{2}x^{2}-2abx + b^{2}\). For the first term \(9x^{2}=(3x)^{2}\), so \(a = 3\) (since \(a^{2}=9\), then \(a=\sqrt{9} = 3\)).

Step2: Identify the constant term

The constant term is \(49\), and \(49 = 7^{2}\), so \(b = 7\) (since \(b^{2}=49\), then \(b=\sqrt{49}=7\)).

Step3: Verify the middle term

Now, check the middle term of the perfect square formula. The middle term should be \(-2abx\). Substituting \(a = 3\) and \(b = 7\), we get \(-2\times3\times7x=-42x\), which matches the middle term of the given trinomial. So, \(9x^{2}-42x + 49=(3x - 7)^{2}\).

Answer:

The first box (for the coefficient of \(x\)) is \(3\) and the second box (for the constant) is \(7\). So the factored form is \((3x - 7)^{2}\), with the first blank being \(3\) and the second blank being \(7\).