Sure! Let's solve this step by step.
### a) Show that [tex]\( f(-1) = 0 \)[/tex]
To verify that [tex]\( f(-1) = 0 \)[/tex], we simply substitute [tex]\(-1\)[/tex] into the function [tex]\( f(x) \)[/tex] and calculate the result.
Given the function:
[tex]\[ f(x) = 6x^3 + 13x^2 + 2x - 5 \][/tex]
Substitute [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 6(-1)^3 + 13(-1)^2 + 2(-1) - 5 \][/tex]
Now, calculate each term:
- [tex]\( 6(-1)^3 = 6(-1) = -6 \)[/tex]
- [tex]\( 13(-1)^2 = 13(1) = 13 \)[/tex]
- [tex]\( 2(-1) = -2 \)[/tex]
- And the constant term is [tex]\(-5 \)[/tex]
Summing these values together:
[tex]\[ f(-1) = -6 + 13 - 2 - 5 \][/tex]
[tex]\[ f(-1) = (13 - 6) - (2 + 5) \][/tex]
[tex]\[ f(-1) = 7 - 7 \][/tex]
[tex]\[ f(-1) = 0 \][/tex]
Hence, it has been shown that [tex]\( f(-1) = 0 \)[/tex].
### b) Find the factors of [tex]\( f(x) \)[/tex]
To find the factors of the polynomial [tex]\( f(x) = 6x^3 + 13x^2 + 2x - 5 \)[/tex], we look for polynomial expressions [tex]\( (x+a)(x+b)(x+c) \)[/tex] that, when multiplied, yield the original polynomial.
Given are the factors:
[tex]\[ f(x) = 6x^3 + 13x^2 + 2x - 5 = (x + 1)(2x - 1)(3x + 5) \][/tex]
Let's verify by expanding these factors to see if they yield the original polynomial.
First, multiply [tex]\( (x + 1) \)[/tex] and [tex]\( (2x - 1) \)[/tex]:
[tex]\[ (x + 1)(2x - 1) = x(2x - 1) + 1(2x - 1) \][/tex]
[tex]\[ = 2x^2 - x + 2x - 1 \][/tex]
[tex]\[ = 2x^2 + x - 1 \][/tex]
Next, multiply this result by [tex]\( (3x + 5) \)[/tex]:
[tex]\[ (2x^2 + x - 1)(3x + 5) \][/tex]
[tex]\[ = 2x^2(3x + 5) + x(3x + 5) - 1(3x + 5) \][/tex]
[tex]\[ = 6x^3 + 10x^2 + 3x^2 + 5x - 3x - 5 \][/tex]
[tex]\[ = 6x^3 + 13x^2 + 2x - 5 \][/tex]
Which is indeed the original polynomial.
Therefore, the factors of the polynomial are correctly given by:
[tex]\[ f(x) = (x + 1)(2x - 1)(3x + 5) \][/tex]
