Question

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

Answer 1

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 \( 8x^2 \):

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

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

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

3. Subtract this from the original polynomial:

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

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

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

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

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

6. 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:

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