Question

which statement about $4x^2 + 19x - 5$ is true? one of the factors is $(4x - 5)$. one of the factors is $(4x + 1)$. one of the factors is $(x + 5)$. one of the factors is $(x - 4)$.

Explanation:

Step1: Factor the quadratic

To factor \(4x^2 + 19x - 5\), we look for two numbers that multiply to \(4\times(-5)= -20\) and add to \(19\). The numbers are \(20\) and \(-1\).
Rewrite the middle term: \(4x^2 + 20x - x - 5\)
Group the terms: \((4x^2 + 20x) + (-x - 5)\)
Factor out the GCF from each group: \(4x(x + 5) - 1(x + 5)\)
Factor out \((x + 5)\): \((4x - 1)(x + 5)\) Wait, no, let's check again. Wait, when we have \(4x^2 + 19x - 5\), let's use the AC method correctly.
\(a = 4\), \(b = 19\), \(c = -5\)
We need two numbers that multiply to \(4\times(-5)= -20\) and add to \(19\). The numbers are \(20\) and \(-1\) (since \(20\times(-1)= -20\) and \(20 + (-1)=19\))
So, rewrite the middle term: \(4x^2 + 20x - x - 5\)
Now, group: \((4x^2 + 20x) + (-x - 5)\)
Factor each group: \(4x(x + 5) - 1(x + 5)\)
Now, factor out \((x + 5)\): \((4x - 1)(x + 5)\)? Wait, that doesn't seem right. Wait, maybe I made a mistake. Let's try another way. Let's use trial and error for factoring.
We need two binomials \((ax + b)(cx + d)\) where \(ac = 4\), \(bd = -5\), and \(ad + bc = 19\)
Possible values for \(ac = 4\): \(a = 4\), \(c = 1\) or \(a = 2\), \(c = 2\) or \(a = 1\), \(c = 4\)
Possible values for \(bd = -5\): \(b = 5\), \(d = -1\) or \(b = -5\), \(d = 1\) or \(b = 1\), \(d = -5\) or \(b = -1\), \(d = 5\)
Let's try \(a = 4\), \(c = 1\), \(b = 5\), \(d = -1\): \((4x + 5)(x - 1)\). Let's multiply: \(4x^2 - 4x + 5x - 5 = 4x^2 + x - 5\). Not correct.
Try \(a = 4\), \(c = 1\), \(b = -5\), \(d = 1\): \((4x - 5)(x + 1)\). Multiply: \(4x^2 + 4x - 5x - 5 = 4x^2 - x - 5\). Not correct.
Try \(a = 4\), \(c = 1\), \(b = 1\), \(d = -5\): \((4x + 1)(x - 5)\). Multiply: \(4x^2 - 20x + x - 5 = 4x^2 - 19x - 5\). Not correct.
Try \(a = 4\), \(c = 1\), \(b = -1\), \(d = 5\): \((4x - 1)(x + 5)\). Multiply: \(4x^2 + 20x - x - 5 = 4x^2 + 19x - 5\). Yes! So the factors are \((4x - 1)(x + 5)\)
Wait, but the options are \((4x - 5)\), \((4x + 1)\), \((x + 5)\), \((x - 4)\)
From our factoring, we have \((4x - 1)(x + 5)\), so one of the factors is \((x + 5)\)
Let's check each option:

• Option 1: \((4x - 5)\). Let's multiply \((4x - 5)(x + a)\) to see if we get \(4x^2 + 19x - 5\). \((4x - 5)(x + a)=4x^2 + (4a - 5)x - 5a\). We need \(4a - 5 = 19\) and \(-5a = -5\). From \(-5a = -5\), \(a = 1\). Then \(4(1) - 5 = -1

eq19\). So not a factor.

• Option 2: \((4x + 1)\). Multiply \((4x + 1)(x + a)=4x^2 + (4a + 1)x + a\). We need \(4a + 1 = 19\) (so \(4a = 18\), \(a = 4.5\)) and \(a = -5\). Contradiction. So not a factor.
• Option 3: \((x + 5)\). From our factoring, \((4x - 1)(x + 5)=4x^2 + 20x - x - 5 = 4x^2 + 19x - 5\). So this is a factor.
• Option 4: \((x - 4)\). Multiply \((x - 4)(4x + a)=4x^2 + (a - 16)x - 4a\). We need \(a - 16 = 19\) (so \(a = 35\)) and \(-4a = -5\) (so \(a = 5/4\)). Contradiction. So not a factor.

Step2: Confirm the correct factor

From the factoring, we have \((4x - 1)(x + 5)=4x^2 + 19x - 5\), so \((x + 5)\) is a factor.

Answer:

One of the factors is \((x + 5)\).