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\((2x - 1)(x - 5)\) ### Problem 2: Factor \(5x^2 - 9x - 2\)

分类: algebra 更新时间: 2026-02-06

题目

Turn 1 Question
  1. $2x^2 - 11x + 5$
  2. $5x^2 - 9x - 2$
  3. $2x^2 + 13x + 7$
  4. $3x^2 + 5x + 2$
  5. $5x^2 - 6x + 1$
  6. $5x^2 + 9x - 2$
Turn 2 Question

For number 3 it’s a -7 not +7

Turn 3 Question

Yes😁

解题步骤

  1. Understand the question
    Turn 1 Question
    1. $2x^2 - 11x + 5$
    2. $5x^2 - 9x - 2$
    3. $2x^2 + 13x + 7$
    4. $3x^2 + 5x + 2$
    5. $5x^2 - 6x + 1$
    6. $5x^2 + 9x - 2$
    Turn 2 Question

    For number 3 it’s a -7 not +7

    Turn 3 Question

    Yes😁

  2. Response
    Turn 1 Answer
    Problem 1: Factor \(2x^2 - 11x + 5\)
  3. Explanation

    Step 1: Find \(ac\) and split middle term

    For \(ax^2+bx+c\), \(a = 2\), \(b=-11\), \(c = 5\). \(ac=2\times5 = 10\). Find two numbers that multiply to \(10\) and add to \(-11\): \(-10\) and \(-1\).
    Rewrite the middle term: \(2x^2-10x - x + 5\)

    Step 2: Group and factor

    Group first two and last two terms: \((2x^2 - 10x)+(-x + 5)\)
    Factor out GCF from each group: \(2x(x - 5)-1(x - 5)\)
    Factor out \((x - 5)\): \((2x - 1)(x - 5)\)

  4. Explanation

    Step 1: Find \(ac\) and split middle term

    \(a = 5\), \(b=-9\), \(c=-2\). \(ac=5\times(-2)=-10\). Find two numbers that multiply to \(-10\) and add to \(-9\): \(-10\) and \(1\).
    Rewrite the middle term: \(5x^2-10x+x - 2\)

    Step 2: Group and factor

    Group: \((5x^2 - 10x)+(x - 2)\)
    Factor out GCF: \(5x(x - 2)+1(x - 2)\)
    Factor out \((x - 2)\): \((5x + 1)(x - 2)\)

  5. Explanation

    Step 1: Find \(ac\) and split middle term

    \(a = 2\), \(b = 13\), \(c = 7\). \(ac=2\times7 = 14\). Find two numbers that multiply to \(14\) and add to \(13\): \(14\) and \(1\).
    Rewrite the middle term: \(2x^2+14x+x + 7\)

    Step 2: Group and factor

    Group: \((2x^2 + 14x)+(x + 7)\)
    Factor out GCF: \(2x(x + 7)+1(x + 7)\)
    Factor out \((x + 7)\): \((2x + 1)(x + 7)\)

  6. Final answer

    \((2x - 1)(x - 5)\)

    Problem 2: Factor \(5x^2 - 9x - 2\)

答案

Response

Turn 1 Answer
Problem 1: Factor \(2x^2 - 11x + 5\)

Explanation

Step 1: Find \(ac\) and split middle term

For \(ax^2+bx+c\), \(a = 2\), \(b=-11\), \(c = 5\). \(ac=2\times5 = 10\). Find two numbers that multiply to \(10\) and add to \(-11\): \(-10\) and \(-1\).
Rewrite the middle term: \(2x^2-10x - x + 5\)

Step 2: Group and factor

Group first two and last two terms: \((2x^2 - 10x)+(-x + 5)\)
Factor out GCF from each group: \(2x(x - 5)-1(x - 5)\)
Factor out \((x - 5)\): \((2x - 1)(x - 5)\)

Answer

\((2x - 1)(x - 5)\)

Problem 2: Factor \(5x^2 - 9x - 2\)

Explanation

Step 1: Find \(ac\) and split middle term

\(a = 5\), \(b=-9\), \(c=-2\). \(ac=5\times(-2)=-10\). Find two numbers that multiply to \(-10\) and add to \(-9\): \(-10\) and \(1\).
Rewrite the middle term: \(5x^2-10x+x - 2\)

Step 2: Group and factor

Group: \((5x^2 - 10x)+(x - 2)\)
Factor out GCF: \(5x(x - 2)+1(x - 2)\)
Factor out \((x - 2)\): \((5x + 1)(x - 2)\)

Answer

\((5x + 1)(x - 2)\)

Problem 3: Factor \(2x^2 + 13x + 7\)

Explanation

Step 1: Find \(ac\) and split middle term

\(a = 2\), \(b = 13\), \(c = 7\). \(ac=2\times7 = 14\). Find two numbers that multiply to \(14\) and add to \(13\): \(14\) and \(1\).
Rewrite the middle term: \(2x^2+14x+x + 7\)

Step 2: Group and factor

Group: \((2x^2 + 14x)+(x + 7)\)
Factor out GCF: \(2x(x + 7)+1(x + 7)\)
Factor out \((x + 7)\): \((2x + 1)(x + 7)\)

Answer

\((2x + 1)(x + 7)\)

Problem 4: Factor \(3x^2 + 5x + 2\)

Explanation

Step 1: Find \(ac\) and split middle term

\(a = 3\), \(b = 5\), \(c = 2\). \(ac=3\times2 = 6\). Find two numbers that multiply to \(6\) and add to \(5\): \(3\) and \(2\).
Rewrite the middle term: \(3x^2+3x+2x + 2\)

Step 2: Group and factor

Group: \((3x^2 + 3x)+(2x + 2)\)
Factor out GCF: \(3x(x + 1)+2(x + 1)\)
Factor out \((x + 1)\): \((3x + 2)(x + 1)\)

Answer

\((3x + 2)(x + 1)\)

Problem 5: Factor \(5x^2 - 6x + 1\)

Explanation

Step 1: Find \(ac\) and split middle term

\(a = 5\), \(b=-6\), \(c = 1\). \(ac=5\times1 = 5\). Find two numbers that multiply to \(5\) and add to \(-6\): \(-5\) and \(-1\).
Rewrite the middle term: \(5x^2-5x - x + 1\)

Step 2: Group and factor

Group: \((5x^2 - 5x)+(-x + 1)\)
Factor out GCF: \(5x(x - 1)-1(x - 1)\)
Factor out \((x - 1)\): \((5x - 1)(x - 1)\)

Answer

\((5x - 1)(x - 1)\)

Problem 6: Factor \(5x^2 + 9x - 2\)

Explanation

Step 1: Find \(ac\) and split middle term

\(a = 5\), \(b = 9\), \(c=-2\). \(ac=5\times(-2)=-10\). Find two numbers that multiply to \(-10\) and add to \(9\): \(10\) and \(-1\).
Rewrite the middle term: \(5x^2+10x - x - 2\)

Step 2: Group and factor

Group: \((5x^2 + 10x)+(-x - 2)\)
Factor out GCF: \(5x(x + 2)-1(x + 2)\)
Factor out \((x + 2)\): \((5x - 1)(x + 2)\)

Answer

\((5x - 1)(x + 2)\)

Turn 2 Answer

Oops, my bad for misreading that! Let me fix problem 3 for you:

Problem 3: Factor \(2x^2 + 13x - 7\)

Explanation

  1. First, calculate \(ac\): \(a=2\), \(c=-7\), so \(ac = 2\times(-7) = -14\)
  2. Find two numbers that multiply to \(-14\) and add up to \(13\): those numbers are \(14\) and \(-1\)
  3. Split the middle term using these numbers: \(2x^2 + 14x - x - 7\)
  4. Group and factor:
  • Group the first two and last two terms: \((2x^2 + 14x) + (-x - 7)\)
  • Factor out the GCF from each group: \(2x(x + 7) - 1(x + 7)\)
  • Factor out the common binomial: \((2x - 1)(x + 7)\)

Answer

\((2x - 1)(x + 7)\)

Did I get that right this time? 😅

Turn 3 Answer

Phew, crisis averted! 😂 Glad I could fix that for you. Want to go over any of these again, or is there something else you feel like chatting about?

Question Analysis

Subject mathematics
Sub Subject algebra
Education Level high school
Difficulty unspecified
Question Type calculation
Multi Question Yes
Question Count 3
Analysis Status completed
Analyzed At 2026-02-06T00:44:53

OCR Text

Show OCR extraction
Yes😁

相关知识点

mathematicsalgebracalculationhigh schoolliked-answerturns-3

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