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Factor Trinomials of the Form ax2+bx+cUsing Trial and Error.

−30q3-140q2−80q

Short Answer

Expert verified

The solution is -10q(q+4)(3q+2).

Step by step solution

01

Step 1. Given information

Consider the trinomial.

−30q3-140q2−80q

02

Step 2. Write the trinomial in descending order and then find the greatest common factor.

The trinomial −30q3-140q2−80qis given in descending order.

There is a greatest common factor in the given trinomial.

Factor the greatest common factor.

role="math" localid="1645100105345" −30q3-140q2−80q=-10q(3q2+14q+8)

03

Step 3. Find the factors of the first term and the last term of the trinomial 3q2+14q+8.

The first term of the trinomial 3q2+14q+8is 3q2.

So, the factors of 3q2are as follows:

3q2=q·3q

The last term of the trinomial q2+14q+8is 8.

Since all the terms in the polynomial q2+14q+8is positive, all the factors of the last term will be positive.

The factors of 8are as follows:

8=1·88=2·4

04

Step 4. Make a table for all the combination of factors of 3q2+14q+8.

Possible factorsProduct
(q+1)(3q+8)
3q2+11q+8
(q+8)(3q+1)
3q2+25q+8
(q+2)(3q+4)
3q2+10q+8
(q+4)(3q+2)
3q2+14q+8

From the table, conclude that the combinationlocalid="1645100413930" (q+4)(3q+2)is correct.

05

Step 5. Check by multiplying all the factors  -10q(q+4)(3q+2).

-10q(q+4)(3q+2)=-10q(q2+2q+12q+8)=-10q(q2+14q+8)=-10q3-140q2-80q

Hence, the factor is -10q(q+4)(3q+2).

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