Which values of x would make a polynomial equal to zero if the factors of the polynomial were (x-2) and (x-11) ?

1 answer:

5 0

I havent done this stuff but I'm assuming x has to be equal to the number its being subtracted from so that the end result would be 0.

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Solve for x: 1 1/5 x -2 1/3 < 1/7 x+ 1 1/2

Firlakuza [10]

Solving inequalities is similar to solving a regular equation. The only thing you need to worry about with inequalities is that when you are dividing or multiplying by a negative number, you must flip the inequality sign. You won't have to worry about that here if the coefficient in front of x is positive when you divide!

Also remember:
- To add/subtract fractions, you have to turn all mixed numbers into improper fractions (if needed), find a common denominator on both fractions (if needed), add/subtract the numerators, put the sum/difference over the common denominator, and simplify if needed.

- To multiply fractions, multiply the numerators and multiply the denominators. Put the product of the numerators over the product of the denominators.

- To divide fractions, remember that dividing by a fraction is the same as multiplying by the inverse of that fraction (aka fraction flipped).

- To turn mixed numbers into improper fractions, multiply the whole number by the denominator of the fraction. Add the numerator of the fraction to the product you get, and put that final sum over the original denominator.

Back to the problem:
You are told that $1 \frac{1}{5} x - 2 \frac{1}{3} \ \textless \ \frac{1}{7} x + 1 \frac{1}{2}$ and you have to solve for x.

1) Using the info for converting mixed numbers to improper fractions from above, you know that $1 \frac{1}{2} = \frac{3}{2}$ and $2 \frac{1}{3} = \frac{7}{3}$ and $1 \frac{1}{5} x = \frac{6}{5} x$ . Now add/subtract using the info above, isolating the variable x by adding $2 \frac{1}{3}$ to both sides and subtracting $\frac{1}{7} x$ from both sides.
$1 \frac{1}{5} x - 2 \frac{1}{3} \ \textless \ \frac{1}{7} x + 1 \frac{1}{2}\\
\frac{6}{5} x - \frac{7}{3} \ \textless \ \frac{1}{7} x + \frac{3}{2}\\
(\frac{6}{5} x - \frac{1}{7} x) \ \textless \ (\frac{3}{2} + \frac{7}{3})\\
(\frac{42}{35} x - \frac{5}{35} x) \ \textless \ (\frac{9}{6} + \frac{14}{6})\\
 \frac{37}{35} x \ \textless \ \frac{23}{6}$

2) Divide both sides by $\frac{37}{35}$ to get the inequality for x. Remember the info for dividing from above:
$\frac{37}{35} x \ \textless \ \frac{23}{6}\\
x \ \textless \ \frac{23}{6} \div \frac{37}{35} \\
x \ \textless \ \frac{23}{6} \times \frac{35}{37} \\
x \ \textless \ \frac{805}{222}$

Your final answer is x < $\frac{805}{222}$ or x < $3 \frac{139}{222}$ .