Question

drag the correct factors to the expression. not all factors will be used. use synthetic division to find all the factors of this polynomial. $4x^3 + 5x^2 - 18x + 9$ $4x - 3$ $x - 3$ $x - 1$ $2x - 3$ $x + 3$ $2x + 1$ ( )( )( ) reset next

Explanation:

Step1: Find possible roots

First, we use the Rational Root Theorem. The possible rational roots of the polynomial \(4x^{3}+5x^{2}-18x + 9\) are factors of \(\frac{9}{4}\), so possible roots are \(\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{9}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{9}{4}\). We test \(x = \frac{3}{4}\) using synthetic division. The coefficient of \(x^{3}\) is \(4\), \(x^{2}\) is \(5\), \(x\) is \(- 18\) and the constant term is \(9\). For synthetic division with root \(a=\frac{3}{4}\), we set up:

$$
LATEXBLOCK0
$$

Since the remainder is \(0\), \((4x - 3)\) is a factor (because if \(x=\frac{3}{4}\) is a root, then \(4x-3 = 0\) implies \(x=\frac{3}{4}\)). After synthetic division, the quotient polynomial is \(4x^{2}+8x - 12\), we can factor out a \(4\) to get \(4(x^{2}+2x - 3)\), and then factor \(x^{2}+2x - 3=(x + 3)(x - 1)\). Wait, no, let's factor \(4x^{2}+8x-12\) directly. Divide by \(4\) first: \(x^{2}+2x - 3=(x + 3)(x - 1)\)? Wait, no, \(x^{2}+2x-3=(x + 3)(x - 1)\) gives \(x^{2}+2x-3\), but \(4x^{2}+8x - 12=4(x^{2}+2x - 3)=4(x + 3)(x - 1)\)? Wait, no, let's check again. Wait, when we did synthetic division with \(x=\frac{3}{4}\), the coefficients after division are \(4,8,-12\), so the quadratic is \(4x^{2}+8x - 12\). Let's factor this quadratic. First, factor out \(4\): \(4(x^{2}+2x - 3)\). Now factor \(x^{2}+2x - 3\): we need two numbers that multiply to \(-3\) and add to \(2\), which are \(3\) and \(- 1\). So \(x^{2}+2x - 3=(x + 3)(x - 1)\). Wait, but we made a mistake. Wait, the original polynomial is \(4x^{3}+5x^{2}-18x + 9\). Wait, maybe we should test \(x = 1\). Let's try \(x = 1\): \(4(1)^{3}+5(1)^{2}-18(1)+9=4 + 5-18 + 9=0\). So \(x = 1\) is a root, so \((x - 1)\) is a factor. Let's do synthetic division with \(x = 1\):

$$
LATEXBLOCK1
$$

The quotient is \(4x^{2}+9x - 9\). Now factor \(4x^{2}+9x - 9\). We need two numbers that multiply to \(4\times(-9)=-36\) and add to \(9\). The numbers are \(12\) and \(- 3\). So \(4x^{2}+12x-3x - 9=4x(x + 3)-3(x + 3)=(4x - 3)(x + 3)\). Wait, so the polynomial \(4x^{3}+5x^{2}-18x + 9=(x - 1)(4x - 3)(x + 3)\)? Wait, no, when we did synthetic division with \(x = 1\), the quotient was \(4x^{2}+9x - 9\), which factors to \((4x - 3)(x + 3)\). So the factors are \((x - 1)\), \((4x - 3)\) and \((x + 3)\). Wait, but let's check the given options. The options are \(4x - 3\), \(x - 3\), \(x - 1\), \(2x - 3\), \(x + 3\), \(2x + 1\). So from our factoring, the factors are \((x - 1)\), \((4x - 3)\) and \((x + 3)\). Let's verify by multiplying \((x - 1)(4x - 3)(x + 3)\). First, multiply \((x - 1)(x + 3)=x^{2}+2x - 3\), then multiply by \((4x - 3)\): \(4x(x^{2}+2x - 3)-3(x^{2}+2x - 3)=4x^{3}+8x^{2}-12x-3x^{2}-6x + 9=4x^{3}+5x^{2}-18x + 9\), which matches the original polynomial. So the factors are \(x - 1\), \(4x - 3\) and \(x + 3\).

Step2: Identify the factors from the options

From the given options, the factors are \(x - 1\), \(4x - 3\) and \(x + 3\). So we need to drag these factors into the boxes.

Answer:

The factors of the polynomial \(4x^{3}+5x^{2}-18x + 9\) are \((x - 1)(4x - 3)(x + 3)\), so the correct factors to drag are \(x - 1\), \(4x - 3\) and \(x + 3\) (among the given options: \(x - 1\), \(4x - 3\), \(x + 3\)).