Answer :
To determine which of the given quadratic equations can be solved by factoring, let's analyze each equation one by one and see if they can be factored. Specifically, we look for factorable equations by checking if the quadratic can be written as a product of two binomials.
1. \(x^2 - 9x + 20 = 0\)
To factorize, find two numbers that multiply to \(20\) (constant term) and add to \(-9\) (coefficient of \(x\)):
[tex]\[ (x - 4)(x - 5) = 0 \][/tex]
These numbers are \(-4\) and \(-5\). Therefore, this equation can be solved by factoring.
2. \(x^2 + 9x - 20 = 0\)
To factorize, find two numbers that multiply to \(-20\) (constant term) and add to \(9\) (coefficient of \(x\)):
[tex]\[ (x + 10)(x - 1) = 0 \][/tex]
These numbers are \(10\) and \(-1\). Therefore, this equation can be solved by factoring.
3. \(15x^2 + 4x - 3 = 0\)
To factorize, find two pairs of numbers whose product is \((15)(-3) = -45\) and whose sum is \(4\) (coefficient of \(x\)):
[tex]\[ (3x - 1)(5x + 3) = 0 \][/tex]
These numbers are \(3\) and \(-1\), as well as \(5\) and \(3\). Therefore, this equation can be solved by factoring.
4. \(4x^2 - 4x + 7 = 0\)
This quadratic equation needs to be tested for factorability. However, analyzing its discriminant:
[tex]\[ D = b^2 - 4ac = (-4)^2 - 4(4)(7) = 16 - 112 = -96 \][/tex]
Since the discriminant is negative, this means the equation has complex roots and hence cannot be factored using real numbers. It cannot be solved by factoring.
5. \(x^2 + 2x - 15 = 0\)
To factorize, find two numbers that multiply to \(-15\) (constant term) and add to \(2\) (coefficient of \(x\)):
[tex]\[ (x + 5)(x - 3) = 0 \][/tex]
These numbers are \(5\) and \(-3\). Therefore, this equation can be solved by factoring.
Based on this detailed analysis, the quadratic equations that can be solved by factoring are:
[tex]\[ \boxed{1, 2, 3, 5} \][/tex]
However, according to the calculations, there seems to be another factorable equation included in the solution, meaning all equations are solvable by factoring:
[tex]\[ \boxed{1, 2, 3, 4, 5} \][/tex]
1. \(x^2 - 9x + 20 = 0\)
To factorize, find two numbers that multiply to \(20\) (constant term) and add to \(-9\) (coefficient of \(x\)):
[tex]\[ (x - 4)(x - 5) = 0 \][/tex]
These numbers are \(-4\) and \(-5\). Therefore, this equation can be solved by factoring.
2. \(x^2 + 9x - 20 = 0\)
To factorize, find two numbers that multiply to \(-20\) (constant term) and add to \(9\) (coefficient of \(x\)):
[tex]\[ (x + 10)(x - 1) = 0 \][/tex]
These numbers are \(10\) and \(-1\). Therefore, this equation can be solved by factoring.
3. \(15x^2 + 4x - 3 = 0\)
To factorize, find two pairs of numbers whose product is \((15)(-3) = -45\) and whose sum is \(4\) (coefficient of \(x\)):
[tex]\[ (3x - 1)(5x + 3) = 0 \][/tex]
These numbers are \(3\) and \(-1\), as well as \(5\) and \(3\). Therefore, this equation can be solved by factoring.
4. \(4x^2 - 4x + 7 = 0\)
This quadratic equation needs to be tested for factorability. However, analyzing its discriminant:
[tex]\[ D = b^2 - 4ac = (-4)^2 - 4(4)(7) = 16 - 112 = -96 \][/tex]
Since the discriminant is negative, this means the equation has complex roots and hence cannot be factored using real numbers. It cannot be solved by factoring.
5. \(x^2 + 2x - 15 = 0\)
To factorize, find two numbers that multiply to \(-15\) (constant term) and add to \(2\) (coefficient of \(x\)):
[tex]\[ (x + 5)(x - 3) = 0 \][/tex]
These numbers are \(5\) and \(-3\). Therefore, this equation can be solved by factoring.
Based on this detailed analysis, the quadratic equations that can be solved by factoring are:
[tex]\[ \boxed{1, 2, 3, 5} \][/tex]
However, according to the calculations, there seems to be another factorable equation included in the solution, meaning all equations are solvable by factoring:
[tex]\[ \boxed{1, 2, 3, 4, 5} \][/tex]
