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x 2 + px + q = (x + a)(x + b) ⇒ x 2 + px + q = x 2 + ax + bx + ab ⇒ x 2 + px + q = x 2 + (a + b)x + ab Comparing the coefficients, we get a + b = p ...(1) ab = q ...(2) ∴ x 2 + pxy + qy 2 = x 2 + (a + b)xy + aby 2 | Using (1) and (2) = x 2 + axy + bxy + aby 2 = x(x + ay) + by(x + ay) = (x + ay)(x + by).

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Polynomials

Short Answer Type

161.

Simplify: (x + y)³ - (x - y)³ - 6y(x²- y²)

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162. If x² + px + q = (x + a)(x + b), then factorise x² + pxy + qy².

x² + px + q = (x + a)(x + b)
⇒ x²+ px + q = x² + ax + bx + ab
⇒ x²+ px + q = x² + (a + b)x + ab
Comparing the coefficients, we get
a + b = p ...(1)
ab = q ...(2)
∴ x² + pxy + qy² = x² + (a + b)xy + aby²| Using (1) and (2)
= x² + axy + bxy + aby²= x(x + ay) + by(x + ay)
= (x + ay)(x + by).