How is finding the factors of a number different from the prime factorization
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1. Using the exponent rule (a^b)·(a^c) = a^(b+c) ...
Simplify. Write in Scientific Notation
2. You know that 256 = 2.56·100 = 2.56·10². After that, we use the same rule for exponents as above.
3. The distributive property is useful for this.
(3x – 1)(5x + 4) = (3x)(5x + 4) – 1(5x + 4)
... = 15x² +12x – 5x –4
... = 15x² +7x -4
4. Look for factors of 8·(-3) = -24 that add to give 2, the x-coefficient.
-24 = -1×24 = -2×12 = -3×8 = -4×6
The last pair of factors adds to give 2. Now we can write
... (8x -4)(8x +6)/8 . . . . . where each of the instances of 8 is an instance of the coefficient of x² in the original expression. Factoring 4 from the first factor and 2 from the second factor gives
... (2x -1)(4x +3) . . . . . the factorization you require