Question
Find whether, or not the first polynomial is a factor of the second: $\frac{3\text{z}^2-13\text{z}+4}{4-\text{z}}$
✓
Answer
$\frac{3\text{z}^2-13\text{z}+4}{4-\text{z}}$
$=\frac{3\text{z}(\text{z}-4)-1(\text{z}-4)}{4-\text{z}}$
$=\frac{(\text{z}-4)(3\text{z}-1)}{4-\text{z}}$
$=\frac{(4-\text{z})(1-3\text{z})}{4-\text{z}}$
$=1-3\text{z}$
Therefore, remainder $= 0$
$(4 - z)$ is a factor of the factor of $3z^2 - 13z + 4$
$=\frac{3\text{z}(\text{z}-4)-1(\text{z}-4)}{4-\text{z}}$
$=\frac{(\text{z}-4)(3\text{z}-1)}{4-\text{z}}$
$=\frac{(4-\text{z})(1-3\text{z})}{4-\text{z}}$
$=1-3\text{z}$
Therefore, remainder $= 0$
$(4 - z)$ is a factor of the factor of $3z^2 - 13z + 4$
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