3a2+30a+63

3a squared+30a+6

3a2+30a+63

3a squared+30a+6

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First, you can factor out the 3 to give 3* (a^2+10a+21). Then, we can use the integral root theorem which states that if a polynomial (x^{n}+a_{n-1}*x^{n-1}+...+a_{1}*x+a_{0}) has real roots, the roots must be factors of a0. Thus given our polynomial, p(a)=a^2+10a+21, a0=21 and our choices for roots are +-1, +-3, +-7, and +-21.

Now, we can eliminate the positive numbers right away from our root choices because all of the terms of the polynomial are positive. Then, we can try -1, -3, -7, and -21. We find that p(-3)=0 and p(-7)=0. Thus our roots are -3 and -7.

Thus, p(a)=(a+3)*(a+7). However, we can not forget the 3 we factored out in the beginning. Putting that back in gives our final result:

3a^2+30a+63 = 3*(a+3)*(a+7)

Notice that this is equivalent to Matthew D's answer.

Factoring trinomials is the opposite of multiplying, so we need to find 2 binomials.

First, factor the 3x^{2}, and place in the front of the binomial:

(3a __)(a __)

Next, to determine the second number in each binomial, list all possible factors of the last term, in this case 63. [what 2 numbers multiplied equal 63]

1 - 63

3 - 21

9 - 7

One of these pairs will finish your binomial. To choose, you could put each option into your binomial and solve to see if the answer results in your original trinomial. Skipping ahead a bit, the pair will 9 & 7. It's important to order the pair appropriately as well.

(3a+7)(a+9) = 3a^{2}+34a+63 *WRONG*

(3a+9)(a+7) = 3a^{2}+30a+63 *CORRECT*

Check your work [FOIL].

(3a+9)(a+7)=3a^{2}+21a+9a+63=3a2+30a+63

~Also, this trinomial uses addition only. If there were a minus sign you'd need to consider that as well. If you get a question with a minus sign, post it and we'll help you through it.

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