Question

1. $x^2 + 5x - 36 = (x - 4)(quad)$
2. $x^2 + 15x + 50 = (x + 5)(quad)$
3. $x^2 - 10x + 21 = (x - 3)(quad)$
4. $x^2 - 11x + 18 = (x - 2)(quad)$
5. $x^2 - 2x - 48 = (x - 8)(quad)$
6. $x^2 + 11x + 24 = (x + 8)(quad)$

Explanation:

Problem 7:

Step1: Recall factoring trinomials. For \(x^2 + bx + c=(x + m)(x + n)\), \(m + n = b\) and \(m\times n = c\). Here, we know one factor is \((x - 4)\), let the other factor be \((x + a)\). Then \((x - 4)(x + a)=x^2+(a - 4)x-4a\). Compare with \(x^2 + 5x - 36\). So \(a - 4 = 5\) and \(-4a=-36\). Solve \(a - 4 = 5\), we get \(a = 9\). Check \(-4\times9=-36\), which matches. So the factor is \((x + 9)\).

Step1: Let the other factor be \((x + a)\). Then \((x + 5)(x + a)=x^2+(a + 5)x+5a\). Compare with \(x^2 + 15x + 50\). So \(a + 5 = 15\) and \(5a = 50\). Solve \(a + 5 = 15\), \(a = 10\). Check \(5\times10 = 50\), correct. So the factor is \((x + 10)\).

Step1: Let the other factor be \((x + a)\). Then \((x - 3)(x + a)=x^2+(a - 3)x-3a\). Compare with \(x^2 - 10x + 21\). So \(a - 3=-10\) and \(-3a = 21\). Solve \(a - 3=-10\), \(a=-7\). Check \(-3\times(-7)=21\), correct. So the factor is \((x - 7)\).

Answer:

\(x + 9\)

Problem 8: