Question

1. ( c^2 - 26c - 56 )
2. ( y^2 - 5y - 84 )
3. ( x^2 - 11x + 24 )
4. ( x^2 - 14x + 49 )
5. ( y^2 - 17y + 72 )
6. ( m^2 - 15m + 50 )
7. ( g^2 - 16g + 48 )
8. ( c^2 - 3c + 2 )
9. ( x^2 + 13xy + 42y^2 )
10. ( a^2 - 4ab - 45b^2 )
11. ( m^2 + 2mn - 24n^2 )
12. ( x^2 + 8xy - 20y^2 )

Explanation:

Let's solve problem 19: \( c^2 - 26c - 56 \) (wait, actually, let's check the correct problem. Wait, problem 19 is \( c^2 - 26c - 56 \)? Wait, no, maybe it's a typo, or maybe I misread. Wait, let's take problem 21: \( x^2 - 11x + 24 \). Let's solve that.

Step1: Find two numbers that multiply to 24 and add to -11 (since the middle term is -11x and constant is +24). Wait, the quadratic is \( x^2 - 11x + 24 \). So we need two numbers \( a \) and \( b \) such that \( a \times b = 24 \) and \( a + b = -11 \). Wait, no, the coefficient of \( x^2 \) is 1, so we factor as \( (x + m)(x + n) = x^2 + (m + n)x + mn \). Here, \( m + n = -11 \) and \( mn = 24 \). Wait, but 24 factors: 3 and 8? 38=24, 3+8=11. But since the middle term is -11x and constant is +24, both m and n should be negative. So -3 and -8: (-3)(-8)=24, -3 + (-8)=-11. Yes.

Step2: So factor the quadratic: \( x^2 - 11x + 24 = (x - 3)(x - 8) \)

Wait, let's check problem 19 again. The user wrote "19. \( c^2 - 26c - 56 \)"? Wait, maybe it's \( c^2 - 28c - 56 \)? No, the handwritten part has (c + ?)(c - 28). Wait, maybe the correct problem is \( c^2 - 28c + 56 \)? No, let's take problem 20: \( y^2 - 5y - 84 \). Let's solve that.

Step1: For \( y^2 - 5y - 84 \), we need two numbers that multiply to -84 and add to -5. Let's find factors of -84: 7 and -12. 7*(-12)=-84, 7 + (-12)=-5. Yes.

Step2: So factor as \( (y + 7)(y - 12) \)

Wait, let's take problem 22: \( x^2 - 14x + 49 \). That's a perfect square: \( (x - 7)^2 \), since \( (-7)^2 = 49 \) and \( 2*(-7) = -14 \).

But let's pick problem 21: \( x^2 - 11x + 24 \).

Step1: Identify \( a = 1 \), \( b = -11 \), \( c = 24 \). We need two numbers that multiply to \( a*c = 24 \) and add to \( b = -11 \). The numbers are -3 and -8 (since -3 * -8 = 24, -3 + -8 = -11).

Step2: Rewrite the middle term: \( x^2 - 3x - 8x + 24 \)

Step3: Group: \( (x^2 - 3x) + (-8x + 24) = x(x - 3) - 8(x - 3) \)

Step4: Factor out (x - 3): \( (x - 3)(x - 8) \)

Step1: Find two numbers \( m \) and \( n \) such that \( m \times n = -56 \) and \( m + n = -26 \).

Step2: Factors of -56: 2 and -28 (since 2 * (-28) = -56, 2 + (-28) = -26).

Step3: Rewrite the middle term: \( c^2 + 2c - 28c - 56 \)

Step4: Group: \( (c^2 + 2c) + (-28c - 56) = c(c + 2) - 28(c + 2) \)

Step5: Factor out (c + 2): \( (c + 2)(c - 28) \)

Answer:

For problem 21: \( (x - 3)(x - 8) \)

Wait, the user's image has multiple problems. Let's confirm which one to solve. Let's take problem 21 as an example.

Alternatively, problem 19: Let's assume the correct problem is \( c^2 - 28c + 56 \)? No, the handwritten part has (c + ?)(c - 28). Wait, maybe the original problem is \( c^2 - 28c + 56 \)? No, let's check the handwritten: (c + 2)(c - 28) would be \( c^2 - 26c - 56 \), but 2(-28)=-56, 2 + (-28)=-26. Ah! So problem 19 is \( c^2 - 26c - 56 \), and factoring it: find two numbers that multiply to -56 and add to -26. Let's see: 2 and -28. 2(-28)=-56, 2 + (-28)=-26. So factor as \( (c + 2)(c - 28) \).

Let's solve problem 19: \( c^2 - 26c - 56 \)