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Factoring Trinomials; I’m a little confused on this question ;(

Factoring Trinomials; I’m a little confused on this question ;(-example-1

2 Answers

1 vote

Answer:

(x + 9)(n - 10)

Explanation:

This trinomial is of the form

x^2 + ax + b

There is no number multiplying x^2.

To factor a trinomial of this type, set up two sets of parentheses:

( )( )

Place the variable inside each set of parentheses on the left side.

(x )(x )

Now look for two numbers that multiply to "b" and add to "a."

Place those two numbers inside the two sets of parentheses on the right side. Include + for a positive number and - for a negative number.

Let's work out your problem.

Factor n^2 - n - 90.

We see that there is no number multiplying the 2nd degree term since we have only n^2, not 2n^2 or 3n^2, etc.

We set up two sets of parentheses and place the variable n on the left side of each.

(n )(n )

Now we need two numbers that multiply to "b" and add to "a". Notice above, in x^2 + ax + b that b is the term that is just a number. "a" is the coefficient of the x-term. In your trinomial, n^2 - n - 90, a = -1 and b = -90. We need two numbers that multiply to -90 and add to -1. Since the two numbers must multiply to a negative number, -90, we must have a positive number and a negative number. 9 and -10 do multiply to -90. Also 9 and -10 add to -1. Now just place 9 and -1 in the parentheses on the right side. Include a "+" with the 9 and the negative sign with the 10.

(n + 9)(n - 10)

User MaLio
by
5.1k points
4 votes
(n-10)(n+9)
is the answer.
If you multiply it out you'll get the original trinomial
-10 and +9 add up to -1, the coefficient of the middle term. They also multiply to -90, which is the last term.
User Perusopersonale
by
5.5k points
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