Practice Makes Perfect Algebra - Carolyn Wheater [18]
If the greatest common factor is a term of the polynomial, remember to express that term as the common factor times 1 and have a 1 in the parentheses for that position.
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EXERCISE 7.1
Factor each polynomial.
1. y2 − 15y
2. 3b2 − 6b
3. 32a2b + 40ab
4. 5y2 + 15y + 20
5. x8y4 − x4y7 + x3y5
6. −a2 −a3 + 2a4
7. 25x4 − 50x5 + 125x7
8. 8r2 + 24rt + 16r
9. 16x2y2 − 48x3y
10. 3x2 y + 6xy + 15x2y2
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Factoring x2 + bx + c
A trinomial of the form x2 + bx + c, if it is not prime, can be factored to the product of two binomials. Just as we say a number is a prime number if its only factors are itself and 1, a polynomial is prime if it’s not factorable. If you factor x2 + 5x + 6, it will factor to(x + r)(x + t). The product of r and t will equal the constant term of the trinomial c, and the sum of r and t will produce the middle coefficient b. To factor x2 + 5x + 6, you need to find a pair of integers that add to 5 and multiply to 6. For small numbers, this isn’t difficult: x2 + 5x + 6 =(x + 3)(x + 2). You can check your factors by multiplying, using the FOIL rule.
If the middle term of the trinomial is negative, as in x2 − 8x + 12, the process of factoring is the same, except that the binomials have negative signs. The factors of x2 − 8x + 12 are(x − 6)(x − 2) because −6 ·(−2) = 12 and −6 +(−2) = −8.
When the constant term is negative, however, it’s a signal that one factor is x plus a constant and one is x minus a constant, and that causes a slight change in your search for factors. If the constant term of the trinomial is negative, you still want two integers that multiply to that number, but you need to remember that adding a positive and a negative will seem like subtracting.
To factor x2 − 9x −90, look for factors of −90, which are just factors of 90 but you’ll make one negative. The factors of 90 are 1 × 90, 2 × 45, 3 × 30, 5 × 18, 6 × 15, and 9 × 10. None of those are going to add to 9 or −9, but that’s OK. You’re actually looking for a pair that subtracts to 9, and 6 × 15 fits the bill. Set up(x 6)(x 15) and then think about the result you want. You want to get −9, so you want a negative sign on the larger number, 15, and a positive sign on the smaller one, 6.
x2 − 9x − 90 =(x + 6)(x − 15)
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EXERCISE 7.2
Factor each polynomial. If the polynomial does not factor, write Prime.
1. x2 + 12x + 35
2. x2 + 11x + 28
3. x2 − 8x + 15
4. x2 − 7x + 12
5. x2 + x − 20
6. x2 − 2x − 3
7. x2 − 11x + 18
8. x2 − 9x − 22
9. x2 + 10x − 39
10. x2 + 12x + 32
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Factoring ax2 + bx + c
When you’re asked to factor a trinomial in which the coefficient of the x2 term is a number other than 1, the factors of that x2 coefficient a, as well as the factors of the constant c, affect the middle term, and the factoring process becomes more a case of trial and error.
To get through that trial-and-error process as efficiently as possible, make a list of the factors of the lead coefficient and a list of the factors of the constant. In theory, you want to check all the possibilities, in all possible combinations, to see whether the “Inner” and the “Outer” from the FOIL rule will combine to form the middle term you need. You need to be very organized, going down your lists in order and crossing factors off when you’re sure they don’t work.
To make things a little easier, you can make a chart of the possible products and look for a pair that will add or subtract to the coefficient of the middle term. To factor 6x2 + 5x − 21, first notice that the factors of 6x2 are 1x · 6x or 2x · 3x and factors of 21 are 1 · 21 or 3 · 7. Put the factors of 6x2 down the side of the chart and the factors of 21 across the top. Keep pairs together. Then fill in the boxes with the products of the number at the beginning of the row and the top of the column.
Look at the numbers on the diagonals of small squares. There’s no way you can get a middle term of 5x from 1x and 126x. The other diagonal in that square will give you 6x and 21x, which could