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Which statement about 4x2 + 19x – 5 is true? One of the factors is (x – 4). One of the factors is (4x + 1). One of the factors is (4x – 5). One of the factors is (x + 5).

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[tex]4 x^{2} + 19x - 5 \\ \\ 4 x^{2} + 20x - x - 5 \\ \\ 4x(x + 5)-(x + 5) \\ \\ (x + 5)(4x - 1) [/tex]

The answer is D) one of the factors of (x + 5). Why? 

(x - 4) is not one of the factors. 
(4x + 1) is not one of the factors.
(4x - 5) is not one of the factors.
(x + 5) IS one of the factors.
TaeKwonDoIsDead

Answer:

Only "One of the factors is (x + 5)." is TRUE.


Step-by-step explanation:

  • What makes [tex](x-4)[/tex] equal to 0?

[tex]x-4=0\\x=4[/tex]

  • What makes [tex](4x+1)[/tex] equal to 0?

[tex]4x+1=0\\4x=-1\\x=-\frac{1}{4}[/tex]

  • What makes [tex](4x-5)[/tex] equal to 0?

[tex]4x-5=0\\4x=5\\x=\frac{5}{4}[/tex]

  • What makes [tex](x+5)[/tex] equal to 0?

[tex]x+5=0\\x=-5[/tex]


Note: Putting all those 4 values into the function [tex]4x^{2}+19x-5[/tex], whichever value makes the function equal to 0, the corresponding expression would be considered a factor of the function [tex]4x^2+19x-5[/tex].

  • Putting [tex]x=4[/tex] into [tex]4x^{2}+19x-5[/tex] gives us:

[tex]4(4)^{2}+19(4)-5\\=135[/tex]

  • Putting [tex]x=-\frac{1}{4}[/tex] into [tex]4x^{2}+19x-5[/tex] gives us:

[tex]4(-\frac{1}{4})^{2}+19(-\frac{1}{4})-5\\=-\frac{19}{2}[/tex]

  • Putting [tex]x=\frac{5}{4}[/tex] into [tex]4x^{2}+19x-5[/tex] gives us:

[tex]4(\frac{5}{4})^{2}+19(\frac{5}{4})-5\\=25[/tex]

  • Putting [tex]x=-5[/tex] into [tex]4x^{2}+19x-5[/tex] gives us:

[tex]4(-5)^{2}+19(-5)-5\\=0[/tex]


As we can see, only putting [tex]x=-5[/tex] in the function gives us a value of 0. So the corresponding expression [tex](x+5)[/tex] is a factor of the function.

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