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 )