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question 1-5 factor completely: $7y + 2xy - 10x - 35$ - $(y - 5)(2x + 7…

Question

question 1-5
factor completely: $7y + 2xy - 10x - 35$

  • $(y - 5)(2x + 7)$
  • $y(7 + 2x) + 5(-2x - 7)$
  • $y(7 - 3x) - 5(3x - 5)$
  • not factorable

Explanation:

Step1: Group the terms

We have the expression \(7y + 2xy-10x - 35\). Group the first two terms and the last two terms: \((7y + 2xy)+(- 10x - 35)\)

Step2: Factor out the GCF from each group

From the first group \(7y + 2xy\), we can factor out \(y\) to get \(y(7 + 2x)\). From the second group \(-10x - 35\), we can factor out \(-5\) to get \(-5(2x + 7)\)
So the expression becomes \(y(2x + 7)-5(2x + 7)\)

Step3: Factor out the common binomial factor

Now we can factor out \((2x + 7)\) from both terms: \((2x + 7)(y - 5)\) which can be rewritten as \((y - 5)(2x + 7)\)

Answer:

\((y - 5)(2x + 7)\) (corresponding to the first option: \(\boldsymbol{(y - 5)(2x + 7)}\))