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.
Hello, kwtolley!
Obviously, there is a typo in the problem . . .
The left side is a quadraticGiven that $\displaystyle (3x-a)(x-2) \:= \:3x^3-32x^2+81x-70$, determine the value of $\displaystyle a.$
. . and cannot possibly equal the cubic on the right.
Hello, kwtolley!
You blew it again: that's 87x . . .
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: .$\displaystyle a = 5$