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lesson 1.3 practice b for use with the lesson \solve x² + bx + c = 0 by factoring\ factor the expression. if the expression cannot be factored, say so. 1. x² + 4x - 21 2. x² - 6x + 5 3. x² + 6x + 8 4. x² - x - 6 5. x² - x - 12 6. x² - 2x - 8 7. x² - 9x + 20 8. x² + 3x - 18 9. x² - 9 10. x² + 8x + 16 11. x² - 11x + 28 12. x² - 2x + 2 prime 13. x² + 4x - 32 14. x² - 3x - 10 15. x² - 25 16. x² - 9x + 14 17. x² - 100 18. x² - 8x - 15 prime solve the equation. 19. x² + x - 6 = 0 20. x² + 3x - 10 = 0 21. x² - 5x + 6 = 0 22. x² - 4x + 4 = 0 23. x² + 7x + 12 = 0 24. x² - 3x - 28 = 0 25. x² - 36 = 0 26. x² - 2x - 15 = 0 27. x² - 11x + 18 = 0 28. 3x² = 48 29. x² - 7x - 4 = -10 30. 9x - 8 = x² find the zeros of the function by rewriting the function in intercept form. 31. y = x² + 8x + 15 32. y = x² - 12x + 32 33. f(x) = x² - 2x - 35 34. y = x² - x - 30 35. g(x) = x² + 10x + 9 36. y = x² - 6x 37. h(x) = x² - 12x + 27 38. y = x² - 9 39. y = x² + 16x + 64
Step1: Factorize \(x^{2}+4x - 21\)
We need to find two numbers that multiply to \(-21\) and add up to \(4\). The numbers are \(7\) and \(- 3\). So \(x^{2}+4x - 21=(x + 7)(x-3)\)
Step2: Factorize \(x^{2}-6x + 5\)
Find two numbers that multiply to \(5\) and add up to \(-6\). The numbers are \(-5\) and \(-1\). So \(x^{2}-6x + 5=(x - 5)(x - 1)\)
Step3: Factorize \(x^{2}+6x + 8\)
Find two numbers that multiply to \(8\) and add up to \(6\). The numbers are \(2\) and \(4\). So \(x^{2}+6x + 8=(x + 2)(x + 4)\)
Step4: Factorize \(x^{2}-x - 6\)
Find two numbers that multiply to \(-6\) and add up to \(-1\). The numbers are \(-3\) and \(2\). So \(x^{2}-x - 6=(x-3)(x + 2)\)
Step5: Factorize \(x^{2}-x - 12\)
Find two numbers that multiply to \(-12\) and add up to \(-1\). The numbers are \(-4\) and \(3\). So \(x^{2}-x - 12=(x-4)(x + 3)\)
Step6: Factorize \(x^{2}-2x - 8\)
Find two numbers that multiply to \(-8\) and add up to \(-2\). The numbers are \(-4\) and \(2\). So \(x^{2}-2x - 8=(x-4)(x + 2)\)
Step7: Factorize \(x^{2}-9x + 20\)
Find two numbers that multiply to \(20\) and add up to \(-9\). The numbers are \(-5\) and \(-4\). So \(x^{2}-9x + 20=(x-5)(x - 4)\)
Step8: Factorize \(x^{2}+3x - 18\)
Find two numbers that multiply to \(-18\) and add up to \(3\). The numbers are \(6\) and \(-3\). So \(x^{2}+3x - 18=(x + 6)(x-3)\)
Step9: Factorize \(x^{2}-9\)
Using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), here \(a=x\) and \(b = 3\), so \(x^{2}-9=(x + 3)(x-3)\)
Step10: Factorize \(x^{2}+8x + 16\)
Using the perfect - square formula \(a^{2}+2ab + b^{2}=(a + b)^{2}\), here \(a=x\) and \(b = 4\), so \(x^{2}+8x + 16=(x + 4)^{2}\)
Step11: Factorize \(x^{2}-11x + 28\)
Find two numbers that multiply to \(28\) and add up to \(-11\). The numbers are \(-7\) and \(-4\). So \(x^{2}-11x + 28=(x-7)(x - 4)\)
Step12: \(x^{2}-2x + 2\) is prime as there are no two real numbers that multiply to \(2\) and add up to \(-2\)
Step13: Factorize \(x^{2}+4x - 32\)
Find two numbers that multiply to \(-32\) and add up to \(4\). The numbers are \(8\) and \(-4\). So \(x^{2}+4x - 32=(x + 8)(x-4)\)
Step14: Factorize \(x^{2}-3x - 10\)
Find two numbers that multiply to \(-10\) and add up to \(-3\). The numbers are \(-5\) and \(2\). So \(x^{2}-3x - 10=(x-5)(x + 2)\)
Step15: Factorize \(x^{2}-25\)
Using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), here \(a=x\) and \(b = 5\), so \(x^{2}-25=(x + 5)(x-5)\)
Step16: Factorize \(x^{2}-9x + 14\)
Find two numbers that multiply to \(14\) and add up to \(-9\). The numbers are \(-7\) and \(-2\). So \(x^{2}-9x + 14=(x-7)(x - 2)\)
Step17: Factorize \(x^{2}-100\)
Using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), here \(a=x\) and \(b = 10\), so \(x^{2}-100=(x + 10)(x-10)\)
Step18: \(x^{2}-8x - 15\) is prime as there are no two real numbers that multiply to \(-15\) and add up to \(-8\)
For solving equations:
Step19: Solve \(x^{2}+x - 6=0\)
Factor the left - hand side: \((x + 3)(x-2)=0\). Then \(x+3 = 0\) or \(x - 2=0\), so \(x=-3\) or \(x = 2\)
Step20: Solve \(x^{2}+3x - 10=0\)
Factor the left - hand side: \((x + 5)(x-2)=0\). Then \(x+5 = 0\) or \(x - 2=0\), so \(x=-5\) or \(x = 2\)
Step21: Solve \(x^{2}-5x + 6=0\)
Factor the left - hand side: \((x-2)(x - 3)=0\). Then \(x-2 = 0\) or \(x - 3=0\), so \(x=2\) or \(x = 3\)
Step22: Solve \(x^{2}-4x + 4=0\)
Factor the left - hand side: \((x - 2)^{2}=0\), so \(x = 2\)
Step23: Solve \(x^{2}+7x + 12=0\)
Factor the left - hand side: \((x + 3)(x + 4)=0\). Then \(x+3 = 0\)…
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