Factor. 9d^2 - 6d + 1 ______

Identify Perfect Square Trinomial: Determine if the quadratic can be factored using the perfect square trinomial formula. A perfect square trinomial is in the form of (a^2 ± 2ab + b^2) which factors to (a ± b)^2. We can check if 9d^2 - 6d + 1 is a perfect square trinomial by identifying a^2, 2ab, and b^2 in the expression. 9d^2 is a perfect square since (3d)^2 = 9d^2. 1 is a perfect square since (1)^2 = 1. The middle term, -6d, should be equal to 2ab. We have a = 3d and b = 1, so 2ab = 2 * 3d * 1 = 6d. Since the middle term is -6d, it matches the form of -2ab.Check for Perfect Square Trinomial: Factor the quadratic using the perfect square trinomial formula. Since we have identified that 9d^2 - 6d + 1 is a perfect square trinomial, we can factor it as (a - b)^2 where a = 3d and b = 1. Therefore, the factored form is (3d - 1)^2.

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Algebra 1

Factor quadratics: special cases

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Q. Factor.

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9d^2 - 6d + 1

Identify Perfect Square Trinomial: Determine if the quadratic can be factored using the perfect square trinomial formula. $\newline$ A perfect square trinomial is in the form of $(a^2 \pm 2ab + b^2)$ which factors to $(a \pm b)^2$ . $\newline$ We can check if $9d^2 - 6d + 1$ is a perfect square trinomial by identifying $a^2$ , $2ab$ , and $b^2$ in the expression. $\newline$ $9d^2$ is a perfect square since $(3d)^2 = 9d^2$ . $\newline$ $1$ is a perfect square since $(1)^2 = 1$ . $\newline$ The middle term, $(a \pm b)^2$ $0$ , should be equal to $2ab$ . We have $(a \pm b)^2$ $2$ and $(a \pm b)^2$ $3$ , so $(a \pm b)^2$ $4$ . $\newline$ Since the middle term is $(a \pm b)^2$ $0$ , it matches the form of $(a \pm b)^2$ $6$ .
Check for Perfect Square Trinomial: Factor the quadratic using the perfect square trinomial formula. $\newline$ Since we have identified that $9d^2 - 6d + 1$ is a perfect square trinomial, we can factor it as $(a - b)^2$ where $a = 3d$ and $b = 1$ . $\newline$ Therefore, the factored form is $(3d - 1)^2$ .