All terms have even powers, factorable to the form (a+b)(a-b)
Difference
There is a formula for the "difference of squares." In this case, the answer is (5x - 6)(5x + 6)
Oh, dude, it's like this: all quadratic equations are polynomials, but not all polynomials are quadratic equations. A quadratic equation is a specific type of polynomial that has a degree of 2, meaning it has a highest power of x^2. So, like, all squares are rectangles, but not all rectangles are squares, you know what I mean?
100a2 - 49b2 is the difference of two perfect squares. Therefore, it is equal to the sum times the difference of the roots: (10a + 7b)(10a - 7b)
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"Difference" implies subtraction. Example: The difference of 8 and 5 is 3 because 8 - 5 = 3. To determine if a polynomial is the difference you probably have to subtract one polynomial from another and check if your answer matches a given polynomial. To clarify the above, the polynomial should be able to be factorised into two distinct factors. For example x^2 - y^2 = (x + y)(x - y). This is the difference of two squares.
It is x^2 -4 = (x-2)(x+2) when factored and it is the difference of two squares
(F-G)(F+G) The difference of two squares.
There is a formula for the difference of squares. In this case, the answer is (C + D)(C - D)
Difference
This expression is the difference of squares. It can be factored to (9 - 7n4)(9 + 7n4)
There is a formula for the "difference of squares." In this case, the answer is (5x - 6)(5x + 6)
62
what is the process of writing a expression as a product? is it Factoring, Quadractic equation, perfect Square trinomial or difference of two squares
To factorise a polynomial completely, first look for the greatest common factor (GCF) of the terms and factor it out. Next, apply techniques such as grouping, using the difference of squares, or recognizing special patterns (like trinomials or perfect squares) to break down the remaining polynomial. Continue this process until you can no longer factor, resulting in a product of irreducible factors. Always check your work by expanding the factors to ensure you return to the original polynomial.
actor a GCF from the expression, if possible. Factor a Trinomial, if possible. Factor a Difference Between Two Squares as many times as possible.
(c - 3d - 2)(c - 3d + 2)