# Thread: Determine the value of a

1. ## Determine the value of a

Given that (3x-a)(x-2)=3x^3-32x^2+81x-70, determine the value of a. My answer was 3 but not if it is right. thanks for any help with this one.

2. Hello, kwtolley!

Obviously, there is a typo in the problem . . .

Given that $(3x-a)(x-2) \:= \:3x^3-32x^2+81x-70$, determine the value of $a.$
The left side is a quadratic
. . and cannot possibly equal the cubic on the right.

3. Originally Posted by kwtolley
Given that $(3x-a)(x-2)=3x^3-32x^2+81x-70$, determine the value of a. My answer was 3 but not if it is right. thanks for any help with this one.
It is possible to solve in terms of $x$ but this number would not be a constant, is that what you're looking for or is there a typo in the way you wrote the problem?

4. ## redo of question sorry about that

Given that (3x-a)(x-2)(x-7)=3x^3-32x^2+81x-70, determine the value of a. My answer is 3 is that right. Thanks again for checking with me.

5. Hello, kwtolley!

You blew it again: that's 87x . . .

Given that $(3x-a)(x-2)(x-7) \:=\:3x^3-32x^2+87x-70$, determine the value of $a$.

Multiply out the left side:
. . $3x^3 - (a + 27)x^2 + (9a + 42)x - 14a \;= \;3x^3 - 32x^2 + 87x - 70$

Two polynomials are equal if their corresponding coefficients are equal.

So we have:
$\begin{array}{cccc}3\:=\:3 \\ -(a + 27) \:= \:-32 \\ 9a + 42 \:= \:87 \\ -14a \:= \:-70\end{array}$

And the solution is: . $a = 5$