Larry P.

asked • 07/19/16# Finding Missing Term on Factored Trinomial or Polynomial

How could I find out the missing term in these examples

1) __x

^{2}-9x-9 = (__x+3)(x-3)2) __x

^{2}-19x-20 = (__x+5)(x-4)3) __x

^{2}-17x-5 = (3x-5)(__x+1)Please show me the complete and clear solutions. Thanks in advance.

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## 2 Answers By Expert Tutors

Kenneth S. answered • 07/19/16

Tutor

4.8
(62)
Algebra II EXPERT will help you survive & prosper

### For the first one...suppose the answer is a (coefficient of x squared). Then on the right side, using the **OI** parts of FOIL, you get -3ax + 3x, or 3x(1-a) and we see (from the left side) that 3(1-a) has to equal -9, or 1-a = -3. *Therefore a = 4.*

### This can give you an idea of how to approach the other two problems.

Larry P.

Could you show me the steps visually like

STEP 1) (ax+3)(x-3)

STEP 2)

STEP 3)

.....

again, thanks

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07/19/16

Mack M. answered • 07/19/16

Tutor

New to Wyzant
Math, Economics, Poli-Sci, & Writing

In order to expand a factored polynomial like (aX+c)(bX+d) you use the foil method. For example, you get (aXbX+aXd+bXc+cd), you can then simplify this into something nicer looking. Remember that the constant term is just the constants c and d multiplied by each other, and the x^1 term is aXd+bXc.

In your first example, you have (aX+3)(x-3)=aX^2-9x-9 and you need to figure out what a is. We know that the cd term is (+3*-3), which is 9, so what is the value for a when (aX)(-3)+(X)(3)=-9x? You can divide both sides by X to get (a)(-3)+(3)=-9. Subtract 3 from both sides, you now have (a)(-3)=-12. Divide both sides by -3 and you get a=4.

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Katharyn F.

problem 1)(x+3)(x-3)=x^{2}-9. So we are missing the -9x in the middle. To get this, we can first try adding a coefficient of 2. Then we get 2x^{2}-6x+3x-9. That won't work, but we see now what we need: we need to have -12x+3x=9x. The 3x is a given, as there is no possible coefficient that could change this when multiplying out. However, we can change the coefficient in front of the first x so that we get -12x.(4x+3)(x-3)=4x, and we have our first solution.^{2}-12x+3x-9=4x^{2}-9x-9Problem 2)can be solved similarly. Try multiplying out, and we see we need to have -24x+5x in the middle to get our -19x. To achieve this, we need to add a coefficient of 6 in the blank space.(6x+5)(x-4)=6x.^{2}-24x+5x-20=6x^{2}-19x-20Problem 3)is a little trickier but can be solved similarly. Multiplying out and calling the missing coefficient N, we have 3Nx2-5Nx+3x-5. So -5Nx+3x must be equal to -17x. So N must be 4, given that -20x+3x=-17x.(3x-5)(4x+1)=12x.^{2}-17x-507/19/16