Use synthetic division to divide the polynomials. \( \left(6 x^{2}-19 x+3\right) \div(x-3)=\square \)
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To start, we set up synthetic division for the polynomial \(6x^2 - 19x + 3\) using \(3\) (the zero of \(x - 3\)). We list the coefficients: \(6\) (for \(x^2\)), \(-19\) (for \(x\)), and \(3\) (constant term). 1. Write \(3\) to the left and the coefficients \(6, -19, 3\) to the right. 2. Bring down the \(6\). 3. Multiply \(3\) by \(6\) to get \(18\), then add this to \(-19\) which gives \(-1\). 4. Multiply \(3\) by \(-1\) to get \(-3\) and add this to \(3\) yielding \(0\). Hence, the result of the synthetic division is \(6x - 1\) with a remainder of \(0\). Therefore, \( (6x^2 - 19x + 3) \div (x - 3) = 6x - 1 \).
