Factor $10 x^2+15 x-70$
A. $5(2 x+7)(x-2)$
B. $5(2 x-7)(x-2)$
C. $5(2 x-2)(x-7)$
D. $5(x-7)(2 x+2)$
Answer (1)
The problem requires factoring the quadratic expression 10 x 2 + 15 x − 70 . We first factor out the GCD, 5, resulting in 5 ( 2 x 2 + 3 x − 14 ) . Then, we factor the quadratic 2 x 2 + 3 x − 14 by finding two numbers that multiply to -28 and add to 3, which are 7 and -4. Rewriting the middle term and factoring by grouping gives us 5 ( 2 x + 7 ) ( x − 2 ) . The final factored expression is 5 ( 2 x + 7 ) ( x − 2 ) .
Explanation
Problem Analysis We are given the quadratic expression 10 x 2 + 15 x − 70 and asked to factor it. We are also given four possible factorizations and must choose the correct one.
Factoring out the GCD First, we can factor out the greatest common divisor (GCD) of the coefficients, which is 5. This gives us 5 ( 2 x 2 + 3 x − 14 ) . Now we need to factor the quadratic 2 x 2 + 3 x − 14 .
Finding the right numbers To factor 2 x 2 + 3 x − 14 , we look for two numbers that multiply to 2 × ( − 14 ) = − 28 and add to 3. These numbers are 7 and -4.
Rewriting and Grouping Now we rewrite the middle term using these numbers: 2 x 2 + 7 x − 4 x − 14 . Then we factor by grouping: x ( 2 x + 7 ) − 2 ( 2 x + 7 ) .
Factoring the binomial Factoring out the common binomial factor ( 2 x + 7 ) , we get ( 2 x + 7 ) ( x − 2 ) . Therefore, the factored form of the original expression is 5 ( 2 x + 7 ) ( x − 2 ) .
Final Answer Comparing this to the given options, we see that the correct factorization is 5 ( 2 x + 7 ) ( x − 2 ) .
Examples
Factoring quadratic expressions is a fundamental skill in algebra and is used in many real-world applications. For example, engineers use factoring to design structures and bridges, ensuring stability and optimal use of materials. Similarly, in economics, factoring can help in analyzing cost and revenue functions to determine break-even points and maximize profits. Understanding factoring allows for efficient problem-solving and optimization in various fields.
