2 years ago

The rule is applied to ΔFGH to produce ΔF"G"H".

Mathematics

1 answer:

Andreas93 [3]2 years ago

7 0

Answer:

(4, –1.5)

Step-by-step explanation:

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Rewrite (2x^2+13x+26) / x+4 in the form q(x)+r(x)/b(x) . Then find q(x) and r(x). In the rewritten expression, q(x) is_____and r

likoan [24]

The value of q(x) is $2 x+5$

The value of r(x) is $6$

Explanation:

The given expression is $\frac{2 x^{2}+13 x+26}{x+4}$

We need to rewrite the expression in the form of $q(x)+\frac{r(x)}{b(x)}$

Simplifying the expression, we get,

$\frac{2 x^{2}+8 x+5x+26}{x+4}$

Separating the fractions, we have,

$\frac{2 x^{2}+8 x}{x+4}+\frac{5 x+26}{x+4}$

$2 x+\frac{5 x+26}{x+4}$ -----------(1)

Now, we shall further simplify the term $\frac{5 x+26}{x+4}$ , we get,

$\frac{5 x+26}{x+4}=\frac{5 x+20}{x+4}+\frac{6}{x+4}$

Common out 5 from the numerator, we have,

$\frac{5 x+26}{x+4}=5+\frac{6}{x+4}$

Substituting the value $\frac{5 x+26}{x+4}=5+\frac{6}{x+4}$ in the equation(1), we get,

$2 x+5+\frac{6}{x+1}$

Thus, the expression $\frac{2 x^{2}+13 x+26}{x+4}=2 x+5+\frac{6}{x+1}$ is in the form of $q(x)+\frac{r(x)}{b(x)}$

Hence, we have,

$q(x)=2 x+5$

$r(x)=6$ and

$b(x)=x+4$

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2 years ago