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.

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- Jun 27th 2006, 09:24 AMkwtolleyDetermine 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.

- Jun 27th 2006, 10:45 AMSoroban
Hello, kwtolley!

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

Quote:

Given that $\displaystyle (3x-a)(x-2) \:= \:3x^3-32x^2+81x-70$, determine the value of $\displaystyle a.$

*quadratic*

. . and cannot possibly equal the cubic on the right. - Jun 28th 2006, 02:12 PMQuickQuote:

Originally Posted by**kwtolley**

- Jun 29th 2006, 09:05 AMkwtolleyredo 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.

- Jun 29th 2006, 05:11 PMSoroban
Hello, kwtolley!

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

Quote:

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

Multiply out the left side:

. . $\displaystyle 3x^3 - (a + 27)x^2 + (9a + 42)x - 14a \;= \;3x^3 - 32x^2 + 87x - 70$

Two polynomials are equal if their corresponding coefficients are equal.

So we have:

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

And the solution is: .$\displaystyle a = 5$