Certainly! Let's solve the given expression step-by-step:
We need to evaluate the following expression:
[tex]\[
\frac{x^2+x-30}{x^2+x-42} \cdot \frac{x^2+13x+42}{x^2-x-20} \div \frac{x+7}{x^2+11x+28}
\][/tex]
To simplify, we first rewrite the division as multiplication by the reciprocal:
[tex]\[
\frac{x^2+x-30}{x^2+x-42} \cdot \frac{x^2+13x+42}{x^2-x-20} \cdot \frac{x^2+11x+28}{x+7}
\][/tex]
Let's simplify each polynomial expression by factoring them wherever possible.
1. Factor [tex]\(x^2 + x - 30\)[/tex]:
[tex]\[
x^2 + x - 30 = (x + 6)(x - 5)
\][/tex]
2. Factor [tex]\(x^2 + x - 42\)[/tex]:
[tex]\[
x^2 + x - 42 = (x + 7)(x - 6)
\][/tex]
3. Factor [tex]\(x^2 + 13x + 42\)[/tex]:
[tex]\[
x^2 + 13x + 42 = (x + 6)(x + 7)
\][/tex]
4. Factor [tex]\(x^2 - x - 20\)[/tex]:
[tex]\[
x^2 - x - 20 = (x - 5)(x + 4)
\][/tex]
5. Factor [tex]\(x^2 + 11x + 28\)[/tex]:
[tex]\[
x^2 + 11x + 28 = (x + 7)(x + 4)
\][/tex]
With these factorizations, the expression becomes:
[tex]\[
\frac{(x + 6)(x - 5)}{(x + 7)(x - 6)} \cdot \frac{(x + 6)(x + 7)}{(x - 5)(x + 4)} \cdot \frac{(x + 7)(x + 4)}{x + 7}
\][/tex]
Next, let's simplify the expression by canceling common factors in the numerator and the denominator:
- The factor [tex]\( (x+7) \)[/tex] appears in the numerator and denominator.
- The factor [tex]\( (x+6) \)[/tex] appears in the numerator and denominator.
- The factor [tex]\( (x-5) \)[/tex] appears in the numerator and denominator.
- The factor [tex]\( (x+4) \)[/tex] appears in the numerator and denominator.
After canceling the common factors, the expression simplifies to:
[tex]\[
\frac{(x + 6)(x \cancel{- 5})}{(x + 7)(x - 6)} \cdot \frac{(x \cancel{+ 6})(x \cancel{+ 7})}{(\cancel{x - 5})(x + 4)} \cdot \frac{\cancel{(x + 7)}(x + 4)}{\cancel{x + 7}}
\][/tex]
We are left with:
[tex]\[
\frac{x^2 + 12x + 36}{x - 6}
\][/tex]
Thus, the simplified form of the given expression is:
[tex]\[
\boxed{\frac{x^2 + 12x + 36}{x - 6}}
\][/tex]
