Question

exponents and polynomials
factoring a perfect square trinomial with leading coefficient greater than 1
factor.
$49y^2 - 14y + 1$

Explanation:

Step1: Recall perfect square trinomial formula

A perfect square trinomial is of the form \(a^2 - 2ab + b^2=(a - b)^2\) or \(a^2+2ab + b^2=(a + b)^2\). For the given trinomial \(49y^{2}-14y + 1\), first identify \(a\) and \(b\).

The first term \(49y^{2}=(7y)^{2}\), so \(a = 7y\). The last term \(1 = 1^{2}\), so \(b = 1\).

Step2: Check the middle term

The middle term of the perfect square trinomial formula is \(- 2ab\) (since the sign of the middle term is negative here). Let's calculate \(-2ab\) with \(a = 7y\) and \(b = 1\): \(-2\times(7y)\times1=-14y\), which matches the middle term of the given trinomial.

Step3: Apply the formula

Using the formula \(a^2-2ab + b^2=(a - b)^2\), substitute \(a = 7y\) and \(b = 1\). So \(49y^{2}-14y + 1=(7y - 1)^{2}\).

Answer:

\((7y - 1)^{2}\)