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Factoring trinomials?

+2 votes
4) -x^2-2x+15
#5) -m^2+3m-2
#6) -p^+5p+14
#7) 2w^2+7w+3
#8) 3y^2+5y+2
#9) 2b^2+b-1
#10) 3n^-2
#11) 5a^2+13a-6
asked Feb 1, 2013 in ALGEBRA 1 by linda Scholar
reshown Feb 1, 2013 by bradely

8 Answers

+2 votes

= - x2 - 2x + 15

Now solve the factor method.

Look at the product of the first and last coefficients: (-1)(15) = - 15

We want two factors of - 15 whose sum middle coefficients of -2, and they are 3 and - 5.

= - x2 - 5x + 3x +15

Take out common terms.

= - x(x + 5) + 3(x + 5)

Take out common factors.

= (x + 5)(- x + 3)

= (x + 5)(3 - x).

answered Feb 1, 2013 by richardson Scholar
+2 votes

= - m2 + 3m - 2

Now solve the factor method.

Look at the product of the first and last coefficients: (-1)(-2) = 2

We want two factors of 2 whose sum middle coefficients of 3, and they are 2 and 1.

= - m2 + 2m + 1m - 2

Take out common terms.

= - m(m - 2) +1(m - 2)

Take out common factors.

= (m - 2)(- m + 1)

= (m - 2)(1 - m).

answered Feb 1, 2013 by richardson Scholar
+1 vote

7).  2w2+7w+3

Now solve the factor method.

Look at the product of the first and last coefficients: (2)(3) = 6

We want two factors of 6 whose sum middle coefficients of 7, and they are 6 and 1.

= 2w2 + 6w + 1w + 3

Take out common terms.

= 2w(w + 3) + 1(w + 3).

Take out common factors.

= (w + 3)(2w + 1).

answered Feb 1, 2013 by richardson Scholar
+1 vote

8). 3y2 + 5y + 2

Now solve the factor method.

Look at the product of the first and last coefficients: (3)(2) = 6

We want two factors of 6 whose sum middle coefficients of 5, and they are 3 and 2.

= 3y2 + 3y + 2y + 2

Take out common terms.

= 3y(y + 1) +2(y + 1).

Take out common factors.

= (y + 1)(3y + 2).

answered Feb 1, 2013 by richardson Scholar
+1 vote

9). 2b2 + b - 1

Now solve the factor method.

Look at the product of the first and last coefficients: (2)(-1) = - 2.

We want two factors of - 2 whose sum middle coefficients of 1, and they are 2 and -1.

= 2b2 + 2b -1b - 1

Take out common terms.

= 2b(b + 1) -1(b + 1)

Take out common factors.

= (b + 1)(2b - 1).

answered Feb 1, 2013 by richardson Scholar
+1 vote

10).  3n-2

Recall: The rules for multiplication also include division as multiplication of inverses. We have already indicated that the inverse relationship can be written as negative power, (a-1) = 1 / a.

= 3 (1/n2)

= 3 / n2

It can be written as (1/n) (3/n)

answered Feb 1, 2013 by richardson Scholar
+1 vote

11). 5a2 + 13a - 6

Now solve the factor method.

Look at the product of the first and last coefficients: (5)(-6) = - 30

We want two factors of - 30 whose sum middle coefficients of 13, and they are 15 and - 2.

= 5a2 + 15a - 2a - 6

Take out common terms.

= 5a(a + 3) - 2(a + 3).

Take out common factors.

= (5a - 2)(a + 3).

answered Feb 1, 2013 by richardson Scholar
+1 vote

6).  -p^ + 5p + 14 (it is wrong)

= -p2 + 5p + 14

Now solve the factor method.

Look at the product of the first and last coefficients: (-1)(14) = - 14

We want two factors of - 14 whose sum middle coefficients of 5, and they are 7 and - 2.

= -p2 + 7p - 2p + 14

Take out common terms.

= - p(p - 7) - 2(p - 7)

Take out common factors.

= (p - 7)(-p - 2).

answered Feb 1, 2013 by richardson Scholar

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