16x^3 + 80x^2 + x + 5 / x+5?

(16x³ + 80x² + x + 5) / (x + 5)

first term:

16x³ + 80x² = 16x²(x+5)       ( Distribute 16x² :16x²(x+5)=16x²(x) +16x²(5)= 16x³ + 80x²)

second term :

x+5 = 1 (x+5)                         (Multiplicative identity 1)

Now   (16x³ + 80x² + x + 5)

=16x² (x+5) + 1 (x+5)

=(x+5)(  16x²+1)

So, the solution is      (x+5) (16x²+1)

(16x³ + 80x² + x + 5) / (x + 5)

first term:

16x³ + 80x² = 16x²(x+5)       ( Distribute 16x² :16x²(x+5)=16x²(x) +16x²(5)= 16x³ + 80x²)

second term :

x+5 = 1 (x+5)                         (Multiplicative identity 1)

Now   (16x³ + 80x² + x + 5)/(x +5 )

=(16x² (x+5) + 1 (x+5)) /(x +5)

=(x+5)(  16x²+1) /( x + 5)

So, the solution is       (16x²+1)

+1 vote
Use synthetic division to find .

Step 1 : Write the terms of the dividend so that the degrees of the terms are in descending order. Then write just the coefficients as shown at the right.

Step 2 : Write the constant r of the divisor (x - r ) to the left. In this case, r = -5. Bring the first coefficient, 16, down.

Step 3 : Multiply the first coefficient by r : -5(16) = -80. Write the product under the second coefficient, 80 and add : 80 + (-80) = 0.

Step 4 : Multiply the sum, 0, by r : -5(0) = 0.

Write the product under the next coefficient, 1 and add : 1 + 0 = 1.

Step 5 : Multiply the sum, 1, by r : -5(1) = -5.

Write the product under the next coefficient, 5 and add : 5 + (-5) = 0.

The numbers along the bottom row are the coefficients of the quotient.

Start with the power of x that is one less than the degree of the dividend.

Thus, the quotient is .