Question

fill in the missing values below one at a time to find the quotient when $9x^3 + 5x - 2$ is divided by $3x - 1$.

	$square$
$3x$	$9x^3$
$-1$

answer
attempt 1 out of 2
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Explanation:

Step1: Find the first term of the quotient

To find the first term of the quotient when \(9x^{3}+5x - 2\) is divided by \(3x-1\), we divide the leading term of the dividend \(9x^{3}\) by the leading term of the divisor \(3x\). Using the rule of exponents for division \(\frac{a^{m}}{a^{n}}=a^{m - n}\) and coefficient division \(\frac{9}{3} = 3\) and \(\frac{x^{3}}{x}=x^{2}\), so \(\frac{9x^{3}}{3x}=3x^{2}\). So the first term of the quotient (the top - left box) is \(3x^{2}\).

Step2: Multiply the divisor by the first term of the quotient and fill in the table

We know that the divisor is \(3x-1\) and the first term of the quotient is \(3x^{2}\).

• Multiply \(3x\) by \(3x^{2}\): \(3x\times3x^{2}=9x^{3}\) (which is already given in the table).
• Multiply \(3x\) by the next term of the quotient (let's find the next term). After the first term \(3x^{2}\), we consider the remaining part of the dividend. The dividend is \(9x^{3}+5x - 2\), and we have already accounted for \(9x^{3}\). Now, we multiply \(- 1\) by \(3x^{2}\) to get \(-3x^{2}\).
• Now, we need to find the next term of the quotient. We bring down the next part of the dividend. The remaining part after subtracting \(9x^{3}-3x^{2}\) from \(9x^{3}+5x - 2\) is \((9x^{3}+5x - 2)-(9x^{3}-3x^{2})=3x^{2}+5x - 2\). Now we divide the leading term of \(3x^{2}+5x - 2\) (which is \(3x^{2}\)) by \(3x\). \(\frac{3x^{2}}{3x}=x\). So the second term of the quotient is \(x\).
• Multiply \(3x\) by \(x\) to get \(3x\times x = 3x^{2}\). Then multiply \(-1\) by \(x\) to get \(-x\).
• Now, the remaining part after subtracting \(3x^{2}-x\) from \(3x^{2}+5x - 2\) is \((3x^{2}+5x - 2)-(3x^{2}-x)=6x - 2\). Divide the leading term \(6x\) by \(3x\) to get \(\frac{6x}{3x}=2\). So the third term of the quotient is \(2\).
• Multiply \(3x\) by \(2\) to get \(6x\) and multiply \(-1\) by \(2\) to get \(-2\).

• Filling in the table:
• The top row (quotient) is \(3x^{2},x,2\)
• The middle row (multiplication by \(3x\)): \(9x^{3},3x^{2},6x\)
• The bottom row (multiplication by \(- 1\)): \(-3x^{2},-x,-2\)

Answer:

The first missing value (top - left box) is \(3x^{2}\)

(If we continue to fill the whole table:

• Top row: \(3x^{2}\), \(x\), \(2\)
• Middle row (under \(3x\)): \(9x^{3}\), \(3x^{2}\), \(6x\)
• Bottom row (under \(-1\)): \(-3x^{2}\), \(-x\), \(-2\)

And the quotient is \(3x^{2}+x + 2\) since \((3x - 1)(3x^{2}+x + 2)=9x^{3}+3x^{2}+6x-3x^{2}-x - 2=9x^{3}+5x - 2\))