When using the inspection method, the number you would add to (and subtract from) the constant term of the numerator so the polynomial in the numerator will have (x+ 3) as a factor is

Respuesta :

Question:

Consider the expression [tex]\frac{x^2 + x -10}{x + 3}[/tex]

When using the inspection method the number you would add to (and subtract from) the constant term of the numerator so the polynomial in the numerator will have (x + 3) as a factor is?

Answer:

The constant to add is 4

Step-by-step explanation:

Given

[tex]\frac{x^2 + x -10}{x + 3}[/tex]

First, we need to get an expression that has x + 3 has its factor.

Represent this expression with: [tex](x + 3)(x + k)[/tex]

Expand

[tex]x^2 + 3x + kx + 3k[/tex]

Group like terms

[tex]x^2 + (3 + k)x + 3k[/tex]

Compare the above expression to: [tex]x^2 + x - 10[/tex]

[tex](3 + k)x = x[/tex]

[tex]3k = -10[/tex]

However, we only consider solving [tex](3 + k)x = x[/tex] for k

[tex](3 + k)x = x[/tex]

[tex]3 + k = 1[/tex]

Subtract 3 from both sides

[tex]3 - 3 + k = 1 - 3[/tex]

[tex]k = 1 - 3[/tex]

[tex]k= -2[/tex]

Substitute -2 for k in [tex](x + 3)(x + k)[/tex]

[tex](x + 3)(x + k) = (x + 3)(x -2)[/tex]

[tex](x + 3)(x + k) = x^2 + 3x - 2x - 6[/tex]

[tex](x + 3)(x + k) = x^2 + x - 6[/tex]

So, the expression that has a factor of x + 3 is [tex]x^2 + x - 6[/tex]

To get the constant term to add/subtract, we have:

[tex]Constant = (x^2 + x - 6) - (x^2 + x - 10)[/tex]

Open brackets

[tex]Constant = x^2 + x - 6 - x^2 - x + 10[/tex]

Collect Like Terms

[tex]Constant = x^2 - x^2+ x - x - 6+ 10[/tex]

[tex]Constant = - 6+ 10[/tex]

[tex]Constant = 4[/tex]