Question

please give me answer

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Answer 1

Answer:

Q.3. (i) $(x - 7) times (x^2 - x + 4)$ $=x^3 - 8x^2 + 11x - 28$

(ii) $\frac{-7}{9} ab^2$ x $\frac{6}{21} ab^2$ $=-\frac{2}{9} ab^2$

(iii) $(\frac{2}{5} a^2 + ab^2)$ x $5(a^2 - \frac{1}{2} b^2)$ $= \frac{a}{2} (2a + 5b^2)(2a^2 - b^2)$

Step-by-step explanation:

Q.3. (i) $(x - 7) times (x^2 - x + 4)$

$=x(x^2 - x + 4) - 7(x^2 - x + 4)$

$=x^3 - x^2 + 4x - 7x^2 + 7x - 28$

$=x^3 - 8x^2 + 11x - 28$

(ii) $\frac{-7}{9} ab^2$ x $\frac{6}{21} ab^2$  $=-\frac{2}{9} ab^2$

(iii) $(\frac{2}{5} a^2 + ab^2)$ x $5(a^2 - \frac{1}{2} b^2)$

$= (2a^2 + 5ab^2)(a^2 - \frac{1}{2} b^2)$

$= \frac{1}{2} (2a^2 + 5ab^2)(2a^2 - b^2)$

$= \frac{a}{2} (2a + 5b^2)(2a^2 - b^2)$