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Factoring Trinomials of the Form ax^2 + bx + c; Perfect Squares
We can factor$a^2 b^2$, the difference of two squares, by finding the terms that produce the perfect squares and substituting these quantities into the factorization form.
When using real numbers, there is no factored form for the sum of two squares.
Perfect square trinomials factor as the square of a binomial.
To recognize them, look for whether (1) the first and last terms are perfect squares, and (2) the middle term is divisible by 2, and when halved, equals the product of the terms that when squared produce the first and last terms.
The polynomial $ax^2+bx+c$ can be factored using a variety of methods.
One such method is trial and error .
Ultimately, the trinomial should be factored in the form $(px+q)(rx+s)$, where p, q, r, and s are constants, and x is a variable.
Using trial and error, we can find values for each of the constants, using the FOIL method to determine whether the constants used produce the trinomial $ax^2+bx+c$.
As seen in , FOIL stands for the order in which one multiplies the four terms - first, outside, inside, last.
The order isn't important; however, the acronym is useful to prevent missing a multiplication term or accidentally multiplying the same terms twice.
We know that the product of px and rx must equal ax2.
Additionally, the sum of products $px \ast s$ and $q \ast rx$ must equal bx.
Finally, the product of q and s must equal c.
Some trinomials, known as perfect square trinomials, can be factored into two equal binomials.
Perfect square trinomials always factor as the square of a binomial.
To recognize a perfect square trinomial, look for the following features:
The first and last terms are perfect squares.
The middle term is divisible by 2.
For example, $x^2-10x+25$ can be identified as a perfect square because x2 is the square of x, and 25 is the square of 5.
The middle term (-10x) is divisible by 2 (equalling -5x).
Given that the coefficient of x2 is 1, we know that the factored form will be $(x+a)(x+b)$, where a and b are to-be-determined coefficients.
We need $x \ast b+a \ast x$ to equal -10x, and a*b to equal 25.
Filling in -5 for a and b, we find a plausible solution that reads (x-5)(x-5), or (x-5)2.
This is a perfect square.
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