Multiply the expressions.
[tex]\frac{3 x^2+2 x-21}{-2 x^2-2 x+12} \cdot \frac{2 x^2+25 x+63}{6 x^2+7 x-49}[/tex]
If [tex]$a=1$[/tex], find the values of [tex]$b, c$[/tex], and [tex]$d$[/tex] that make the given expression equivalent to the expression below. [tex]\frac{a x+b}{c x+d}[/tex]
[tex]$b=\square, c=\square, \text { and } d=\square$[/tex]
Answer (1)
Factor the quadratic expressions in the numerators and denominators.
Cancel out the common factors.
Simplify the resulting expression.
Identify the values of b , c , and d in the simplified expression: b = 9 , c = − 2 , d = 4 .
Explanation
Problem Analysis We are given the expression − 2 x 2 − 2 x + 12 3 x 2 + 2 x − 21 ⋅ 6 x 2 + 7 x − 49 2 x 2 + 25 x + 63 and we want to simplify it to the form c x + d a x + b where a = 1 . This means we need to factor the quadratics, cancel common terms, and rewrite the result in the desired form.
Factoring Quadratics First, we factor the quadratic expressions:
3 x 2 + 2 x − 21 = ( 3 x − 7 ) ( x + 3 )
− 2 x 2 − 2 x + 12 = − 2 ( x 2 + x − 6 ) = − 2 ( x + 3 ) ( x − 2 )
2 x 2 + 25 x + 63 = ( 2 x + 7 ) ( x + 9 )
6 x 2 + 7 x − 49 = ( 3 x − 7 ) ( 2 x + 7 )
Substituting Factored Forms Now we substitute the factored forms into the original expression:
− 2 ( x + 3 ) ( x − 2 ) ( 3 x − 7 ) ( x + 3 ) ⋅ ( 3 x − 7 ) ( 2 x + 7 ) ( 2 x + 7 ) ( x + 9 )
Canceling Common Factors Next, we cancel out the common factors:
-2 ( x + 3 ) ( x − 2 ) ( 3 x − 7 ) ( x + 3 ) ⋅ ( 3 x − 7 ) ( 2 x + 7 ) ( 2 x + 7 ) ( x + 9 ) = − 2 ( x − 2 ) x + 9
Simplifying and Comparing Now we simplify the expression:
− 2 ( x − 2 ) x + 9 = − 2 x + 4 x + 9
So we have − 2 x + 4 x + 9 . We want to express this in the form c x + d a x + b where a = 1 . Comparing the expression − 2 x + 4 x + 9 with c x + d a x + b , we see that a = 1 , b = 9 , c = − 2 , and d = 4 .
Final Answer Therefore, the values are b = 9 , c = − 2 , and d = 4 .
Examples
This type of algebraic simplification is useful in many areas of engineering and physics, where complex equations need to be simplified to make calculations easier. For example, in circuit analysis, simplifying transfer functions can help engineers understand the behavior of a circuit more easily. Similarly, in control systems, simplifying the equations of motion can help in designing controllers that stabilize a system.
