What is the third linear factor of the polynomial 16x3+68x2+32x−35 when 4x+5 and 2x−1 are known factors

1 answer

To find the third linear factor of the polynomial \( 16x^3 + 68x^2 + 32x - 35 \) given that \( 4x + 5 \) and \( 2x - 1 \) are known factors, we can use polynomial division.

First, we can multiply the known factors together:

\[ (4x + 5)(2x - 1) = (4x)(2x) + (4x)(-1) + (5)(2x) + (5)(-1) = 8x^2 - 4x + 10x - 5 = 8x^2 + 6x - 5. \]

Now, we will perform polynomial division of the original polynomial \( 16x^3 + 68x^2 + 32x - 35 \) by this product \( 8x^2 + 6x - 5 \).

  1. Divide the leading term of the original polynomial \( 16x^3 \) by the leading term of the divisor \( 8x^2 \):

\[ \frac{16x^3}{8x^2} = 2x. \]

  1. Multiply \( 2x \) by the entire divisor \( 8x^2 + 6x - 5 \):

\[ 2x(8x^2 + 6x - 5) = 16x^3 + 12x^2 - 10x. \]

  1. Subtract this from the original polynomial:

\[ (16x^3 + 68x^2 + 32x - 35) - (16x^3 + 12x^2 - 10x) = (68x^2 - 12x^2) + (32x + 10x) - 35 = 56x^2 + 42x - 35. \]

  1. Now, divide the leading term \( 56x^2 \) by \( 8x^2 \):

\[ \frac{56x^2}{8x^2} = 7. \]

  1. Multiply \( 7 \) by the entire divisor \( 8x^2 + 6x - 5 \):

\[ 7(8x^2 + 6x - 5) = 56x^2 + 42x - 35. \]

  1. Subtract this from \( 56x^2 + 42x - 35 \):

\[ (56x^2 + 42x - 35) - (56x^2 + 42x - 35) = 0. \]

Since the remainder is zero, we conclude that:

\[ 16x^3 + 68x^2 + 32x - 35 = (4x + 5)(2x - 1)(2x + 7). \]

Thus, the third linear factor of the polynomial is:

\[ \boxed{2x + 7}. \]