Question

if $g(x) = 4x^{4} - 23x^{3} + 10x^{2} + 25x$, use synthetic division to find $g(5)$.

Explanation:

Step1: Factor out x from g(x)

First, we factor out \( x \) from the polynomial \( g(x) = 4x^4 - 23x^3 + 10x^2 + 25x \). So, \( g(x)=x(4x^3 - 23x^2 + 10x + 25) \). Now we can use synthetic division on the cubic polynomial \( 4x^3 - 23x^2 + 10x + 25 \) with root \( x = 5 \).

Step2: Set up synthetic division

For synthetic division with root \( 5 \) and coefficients of the cubic polynomial \( 4, -23, 10, 25 \):

Bring down the leading coefficient \( 4 \).

Step3: Multiply and add (first time)

Multiply \( 4 \) by \( 5 \) to get \( 20 \). Then add \( -23 + 20=-3 \).

Step4: Multiply and add (second time)

Multiply \( -3 \) by \( 5 \) to get \( -15 \). Then add \( 10+(-15) = -5 \).

Step5: Multiply and add (third time)

Multiply \( -5 \) by \( 5 \) to get \( -25 \). Then add \( 25+(-25)=0 \).

The result of synthetic division for the cubic polynomial gives a remainder of \( 0 \), and the quotient is \( 4x^2-3x - 5 \). But since \( g(x)=x(4x^3 - 23x^2 + 10x + 25) \), and when we divide \( 4x^3 - 23x^2 + 10x + 25 \) by \( x - 5 \) the remainder is \( 0 \), we can also use the Remainder Theorem which states that the remainder of dividing a polynomial \( f(x) \) by \( x - a \) is \( f(a) \). Alternatively, since we have \( g(x)=x(4x^3 - 23x^2 + 10x + 25) \), and the cubic polynomial evaluated at \( x = 5 \) is \( 0 \) (from synthetic division remainder), then \( g(5)=5\times0 = 0 \)? Wait, no, wait. Wait, let's re - evaluate. Wait, actually, when we do synthetic division for the original polynomial (after factoring out \( x \)), but maybe a better way: Let's consider the original polynomial \( g(x)=4x^4 - 23x^3 + 10x^2 + 25x \). We can write it as \( g(x)=4x^4 - 23x^3 + 10x^2 + 25x+0 \) (adding the constant term \( 0 \) since there is no constant term originally). Now, set up synthetic division for \( x = 5 \) with coefficients \( 4, -23, 10, 25, 0 \):

Bring down \( 4 \).

Multiply \( 4\times5 = 20 \), add to \( -23 \): \( -23 + 20=-3 \)

Multiply \( -3\times5=-15 \), add to \( 10 \): \( 10-15=-5 \)

Multiply \( -5\times5 = -25 \), add to \( 25 \): \( 25-25 = 0 \)

Multiply \( 0\times5=0 \), add to \( 0 \): \( 0 + 0=0 \)

So the remainder when \( g(x) \) is divided by \( x - 5 \) is \( 0 \), which means \( g(5)=0 \).

Wait, let's verify by direct substitution:

\( g(5)=4\times(5)^4-23\times(5)^3 + 10\times(5)^2+25\times(5) \)

\( =4\times625-23\times125 + 10\times25+125 \)

\( =2500-2875 + 250+125 \)

\(=(2500 + 250+125)-2875 \)

\(=2875-2875=0 \)

Answer:

\( 0 \)