9514 1404 393
Answer:
- x(4-x)
- no factorization. x² -8y³
- (x +18)(x -3)
- (2xy +5)(3x -5)
Step-by-step explanation:
Factoring is taking apart products. It helps immensely if you are familiar with how products are formed. That is, it helps to be familiar with your multiplication tables and the rules of divisibility. It also helps to have thorough understanding of the distributive property and the forms produced by the multiplication of binomials.
The first step is to look for any factors that are common to all of the terms you are factoring. Next steps include ...
- look for any special forms (square of binomial, difference of squares, sum or difference of cubes)
- see if factoring by grouping can work.
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1.1.1 4x -x^2
There is only one constant, so no constant factors. The variable x appears in all terms, with a lowest power of 1, so x^1 = x is a factor.
= x(4 -x) . . . . . . . cannot be factored further
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1.1.2 x² -8y³ . . . . we assume this is the intended expression. (It helps to use appropriate (and consistent) exponent presentation: x^2 -8y^3, or x² -8y³.
There are no common numerical factors, no common variable factors, and the form is not the difference of squares or cubes. This expression cannot be factored.
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1.1.3 x² +15x -54
This is a trinomial in a single variable (x), so may be factorable as the product of two binomials. The product of binomial factors will have the form ...
(x -p)(x +q) = x² +(q-p)x -pq
This means we are looking for two factors of 54 that have a difference of 15. Here is where your familiarity with numerical factoring is helpful.
54 = 54×1 = 27×2 = 18×3 = 9×6
Differences of the factors are 54-1=53, 25, 15, 3. The factor pair we're looking for is 18×3. Comparing that to the above form, we see that q=18 and p=3, so the factorization is ...
= (x -3)(x +18)
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1.1.4 6x²y -10xy +15x -25
There are no numerical or variable factors that are seen in all terms. However, we notice that pairs of terms have common factors. Factoring those out gives ...
= (2xy)(3x -5) +5(3x -5)
At this point, we notice that (3x -5) is a factor of both of these terms, so we can factor it out. This gives the complete factorization ...
= (2xy +5)(3x -5)
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Additional comment
Many methods are promoted for factoring trinomials. It is helpful to understand that the product of factors looks like ...
(ax +b)(cx +d) = acx² +(ad+bc)x +bd
That is to say, the product of the x² coefficient and the constant term (ac·bd) will have factors (ad, bc) that have a sum equal to the linear term coefficient. Once you realize this, finding the factorization is not so difficult. You look at the factor pairs of abcd in the same way we looked at factor pairs of -54 in problem 1.1.3 above.
