fill in the missing values below one at a time to find the quotient when $12x^{3}+7x^{2}+28x - 30$ is…

Explanation:

Step1: Find the first term of the quotient

To find the first - term of the quotient, divide the leading term of the dividend (12x^{3}) by the leading term of the divisor (4x). [ \frac{12x^{3}}{4x}=3x^{2} ]

Step2: Fill in the products

• Multiply (3x^{2}) by (- 3): (3x^{2}\times(-3)=-9x^{2})
• The second term of the dividend is (7x^{2}). After subtracting (-9x^{2}) (equivalent to adding (9x^{2})) from (7x^{2}) in the long - division process, we consider the next part. Let the second term of the quotient be (ax). Then (4x\times ax = 4ax^{2}), and (4ax^{2}+(-9x^{2})=7x^{2}), so (4a = 16), (a = 4) (another way: we know that ((3x^{2}+4x + b)(4x - 3)=12x^{3}+7x^{2}+28x-30)).
• Multiply (4x) by (4x): (4x\times4x = 16x^{2})
• Multiply (4x) by (-3): (4x\times(-3)=-12x)
• The third term of the dividend is (28x). Let the third term of the quotient be (b). Then (4x\times b=4bx), and (4bx+(-12x)=28x), so (b = 10)
• Multiply (10) by (4x): (10\times4x = 40x)
• Multiply (10) by (-3): (10\times(-3)=-30)

Answer:

The first box (top - left) is (3x^{2}), the box below (12x^{3}) (left - middle) is (-9x^{2}), the middle - top box is (4x), the middle - middle box is (16x^{2}), the middle - right box is (40x), the top - right box is (10), the right - middle box is (28x) (already given in the dividend structure, but through multiplication (4x\times7x) conceptually related), the bottom - right box is (-30) (matches the dividend's constant term). The quotient is (3x^{2}+4x + 10)