I do not know how to approach this problem but i tried to use synthetic division and I got $k=-4$. Is this the only value or are there more.
Determine all values of k so that $3x+2$ is a factor of the polynomial function $h(x)=6x^3-5x^2-12x+k$
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1https://en.m.wikipedia.org/wiki/Polynomial_remainder_theorem – 2017-01-01
4 Answers
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$3x + 2$ is a factor of the equation.
So, $$3x + 2 = 0$$ $$x = \frac{-2}{3}$$ Put this value in equation.
So we have, $$ 6\left(\frac{-2}{3}\right)^3 - 5\left(\frac{-2}{3}\right)^2 - 12\left(\frac{-2}{3}\right) + k = 0$$ $$ 6 \left(\frac{-8}{27}\right) - 5 \left(\frac{4}{9}\right) + 8 + k = 0$$ $$ \left(\frac{-16}{9}\right) - \left(\frac{20}{9}\right) + 8 + k = 0$$ $$k = \left(\frac{16}{9}\right) + \left(\frac{20}{9}\right) - 8$$ $$k = \frac{ 16 + 20 - 72}{9}$$ $$k = \frac{-36}{9}$$ $$k = -4$$
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Hint:
Substitute $\frac{-2}{3}$ in the given polynomial and equate it to $0$.
Then solve for $k$.
Explanation:
Given $ax+b$ is a factor of some polynomial $\Rightarrow$ $\frac{-b}{a}$ is a root of the polynomial.
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$3x+2$ is a factor of $h(x)$ implies $h(\frac{-2}{3}) =0;$ Thus find the value of $k$
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Alternative Approach:
Since the polynomial is of degree $3$, and notice coefficients of the $x^3$ and constant term,
$$6x^3-5x^2-12x+k=(3x+2)(2x^2+ax+\frac k2)$$
for some constant $a$.
Expand and solve for $a$ and $k$.