QuestionApril 14, 2026

Factor completely. 2j^2-j-10 square

Factor completely. 2j^2-j-10 square
Factor completely.
2j^2-j-10
square

Solution
4.7(294 votes)

Answer

(j+2)(2j - 5) Explanation 1. Multiply leading coefficient and constant 2 \times (-10) = -20 We need two numbers whose product is -20 and sum is -1. 2. Find the pair of numbers The pair is 4 and -5. 3. Split the middle term 2j^{2} - j - 10 = 2j^{2} + 4j - 5j - 10 4. Group terms and factor (2j^{2} + 4j) - (5j + 10) = 2j(j+2) - 5(j+2) 5. Factor common binomial (j+2)(2j - 5)

Explanation

1. Multiply leading coefficient and constant <br /> $2 \times (-10) = -20$ <br />We need two numbers whose product is $-20$ and sum is $-1$.<br /><br />2. Find the pair of numbers <br /> The pair is $4$ and $-5$.<br /><br />3. Split the middle term <br /> $2j^{2} - j - 10 = 2j^{2} + 4j - 5j - 10$<br /><br />4. Group terms and factor <br /> $(2j^{2} + 4j) - (5j + 10)$ <br /> $= 2j(j+2) - 5(j+2)$<br /><br />5. Factor common binomial <br /> $(j+2)(2j - 5)$
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