Answer:
Only "One of the factors is (x + 5)." is TRUE.
Step-by-step explanation:
- What makes \((x-4)\) equal to 0?
\(x-4=0\\x=4\)
- What makes \((4x+1)\) equal to 0?
\(4x+1=0\\4x=-1\\x=-\frac{1}{4}\)
- What makes \((4x-5)\) equal to 0?
\(4x-5=0\\4x=5\\x=\frac{5}{4}\)
- What makes \((x+5)\) equal to 0?
\(x+5=0\\x=-5\)
Note: Putting all those 4 values into the function \(4x^{2}+19x-5\), whichever value makes the function equal to 0, the corresponding expression would be considered a factor of the function \(4x^2+19x-5\).
- Putting \(x=4\) into \(4x^{2}+19x-5\) gives us:
\(4(4)^{2}+19(4)-5\\=135\)
- Putting \(x=-\frac{1}{4}\) into \(4x^{2}+19x-5\) gives us:
\(4(-\frac{1}{4})^{2}+19(-\frac{1}{4})-5\\=-\frac{19}{2}\)
- Putting \(x=\frac{5}{4}\) into \(4x^{2}+19x-5\) gives us:
\(4(\frac{5}{4})^{2}+19(\frac{5}{4})-5\\=25\)
- Putting \(x=-5\) into \(4x^{2}+19x-5\) gives us:
\(4(-5)^{2}+19(-5)-5\\=0\)
As we can see, only putting \(x=-5\) in the function gives us a value of 0. So the corresponding expression \((x+5)\) is a factor of the function.
