Use synthetic substitution to evaluate P(x) = x^3 + 3x^2 – 6 for x = 5

3 answers
Question:

Use synthetic substitution to evaluate P(x) = x^3 + 3x^2 – 6 for x = 5

Answers

[tex]\frac{x^3-3x^2+4x-7}{x-3} =(x^2+4)+\frac{5}{x-3}[/tex]

Step-by-step explanation:

We are given polynomial as

[tex]P(x)=x^3-3x^2+4x-7[/tex]

and it is divided by x=3 or x-3

we can use synthetic division

so, we got

Remainder =5

Quotient is

[tex]=x^2+0x+4[/tex]

[tex]=x^2+4[/tex]

we can write as

[tex]\frac{x^3-3x^2+4x-7}{x-3} =(x^2+4)+\frac{5}{x-3}[/tex]



[tex]Use synthetic substitution to evaluate the polynomial p(x) = x^3-3x^2+4x-7 for x=3[/tex]

When x=4

P(x) =4 - 4(4)^2 + 4(4) -5

=4 - 4(16) + 16 -5

=4 - 64 +11

= -60 + 11

= -49

I'm assuming that is 2x squared, and not 2 multiplied by 2, as a given question usually wouldn't come in such a silly format. I do wish people would start using (2x^2) or (2x²), though.

Just substitute the value of x (-3) in to the equation to find the value of y, or P(x).

2(-3)² + 7(-3) - 1 = -4

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