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Solve These Quadratics by Factorizing pls

Solve These Quadratics by Factorizing pls-example-1
User Ramiramilu
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1 Answer

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(a) x² + 5x + 4 = 0, is factorized as (x + 4)(x + 1) = 0.

(b) x² + 5x + 6 = 0, is factorized as (x + 2)(x + 3) = 0.

(c). x² + 7x + 6 = 0, this is factorized as (x + 6)(x + 1) = 0.

(d) x² + 10x + 16 = 0, this is factorized as (x + 2)(x + 8) = 0

(e), x² + 10x + 21 = 0, this is factorized as (x + 7)(x + 3) = 0.

How to solve the quadratic equation?

The solution of the quadratic equation by factorization method is determined as follows.

(a) x² + 5x + 4 = 0, is factorized as;

x² + 5x + 4 = 0

x² + x + 4x + 4 = 0

x(x + 1) + 4(x + 1) = 0

(x + 4)(x + 1) = 0

(b) x² + 5x + 6 = 0, is factorized as;

x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

(c). x² + 7x + 6 = 0, this is factorized as;

x² + 7x + 6 = 0

x² + x + 6x + 6 = 0

x (x + 1) + 6(x + 1) = 0

(x + 6)(x + 1) = 0

(d) x² + 10x + 16 = 0, this is factorized as;

x² + 10x + 16 = 0

x² + 2x + 8x + 16 = 0

x(x + 2) + 8(x + 2) = 0

(x + 2)(x + 8) = 0

(e), x² + 10x + 21 = 0, this is factorized as;

x² + 10x + 21 = 0

x² + 3x + 7x + 21 = 0

x( x + 3) + 7 (x + 3) = 0

(x + 7)(x + 3) = 0

User Nerfologist
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