Question

What is the third linear factor of the polynomial 16x3+68x2+32x−35%0D%0A16%0D%0A𝑥%0D%0A3%0D%0A+%0D%0A68%0D%0A𝑥%0D%0A2%0D%0A+%0D%0A32%0D%0A𝑥%0D%0A−%0D%0A35%0D%0A when 4x+5%0D%0A4%0D%0A𝑥%0D%0A+%0D%0A5%0D%0A and 2x−1%0D%0A2%0D%0A𝑥%0D%0A−%0D%0A1%0D%0A 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.

1. **Multiply the known factors together:**
\[
(4x + 5)(2x - 1)
\]
Expanding this:
\[
= 4x \cdot 2x + 4x \cdot (-1) + 5 \cdot 2x + 5 \cdot (-1)
\]
\[
= 8x^2 - 4x + 10x - 5
\]
\[
= 8x^2 + 6x - 5
\]

2. **Now, divide the original polynomial by this product:**
We need to divide \( 16x^3 + 68x^2 + 32x - 35 \) by \( 8x^2 + 6x - 5 \).

Using polynomial long division:
- Divide the leading term \( 16x^3 \) by the leading term of the divisor \( 8x^2 \) to get \( 2x \).
- Multiply \( 2x \) by the entire divisor \( 8x^2 + 6x - 5 \):
\[
2x(8x^2 + 6x - 5) = 16x^3 + 12x^2 - 10x
\]
- Subtract this from the original polynomial:
\[
(16x^3 + 68x^2 + 32x - 35) - (16x^3 + 12x^2 - 10x) = (68x^2 - 12x^2) + (32x + 10x) - 35
\]
\[
= 56x^2 + 42x - 35
\]

3. **Now repeat the division process:**
- Divide the leading term \( 56x^2 \) by the leading term of the divisor \( 8x^2 \) to get \( 7 \).
- Multiply \( 7 \) by the entire divisor \( 8x^2 + 6x - 5 \):
\[
7(8x^2 + 6x - 5) = 56x^2 + 42x - 35
\]
- Subtract this from what we just calculated:
\[
(56x^2 + 42x - 35) - (56x^2 + 42x - 35) = 0
\]

Since we have obtained 0 after performing the division, we conclude that the third factor is:
\[
\text{Third factor} = 2x + 7
\]

4. **Final Result:**
Thus, the third linear factor of the polynomial \( 16x^3 + 68x^2 + 32x - 35 \) is \( \boxed{2x + 7} \).