Question

find the factors for any 5 quadratic polynomials​

Answer 1

Answer:

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Answer 2

Answer:

1. x^2 + 5x + 6 = 0

-3 and -2 are the roots of the equation. Verify by substituting the roots in the given equation and check if the value equals 0.

Factor 1: (x + 3)

LHS = x2 + 5x + 6 = (-3)2 + 5 × -3 + 6 = 9 -15 + 6 = 0 = RHS

Factor 2: (x + 2)

LHS = x2 + 5x + 6 = (-2)2 + 5 × -2 + 6 = 4 -10 + 6 = 0 = RHS

Thus the equation has 2 factors (x + 3) and (x + 2)

2. x^2 - 9 = 0

3 and -3 are the two roots of the equation. Verify by substituting the roots in the given equation and check if the value equals 0.

3^2 - 9 = 9 - 9 = 0

(-3)^2 - 9 = 9 - 9 = 0

Thus the equation has 2 factors (x+3) and (x-3)

3. 2(x ^2 + 1) = 5x

Solution:

Expand the equation and move all the terms to the left of the equal sign.

⟹ 2x 2 – 5x + 2 = 0

⟹ 2x 2 – 4x – x + 2 = 0

⟹ 2x (x – 2) – 1(x – 2) = 0

⟹ (x – 2) (2x – 1) = 0

Equate each factor equal to zero and solve

⟹ x – 2 = 0 or 2x – 1 = 0

⟹ x = 2 or x = 1212

Therefore, the solutions are x = 2, 1/2.

4. 3x ^2 – 8x – 3 = 0

Solution

3x^ 2 – 9x + x – 3 = 0

⟹ 3x (x – 3) + 1(x – 3) = 0

⟹ (x – 3) (3x + 1) = 0

⟹ x = 3 or x = -13

5. (2x – 3)^2 = 25  

⟹ 4x 2 – 12x + 9 – 25 = 0

⟹ 4x 2 – 12x – 16 = 0

Divide each term by 4 to get;

⟹ x 2 – 3x – 4 = 0

⟹ (x – 4) (x + 1) = 0

⟹ x = 4 or x = -1

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