Factorise the expressions and divide them as directed.
$39 y^{3} (50 y^{2} - 98) \div 26 y^{2} (5 y + 7)$

3 y (5 y - 7)

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3 y (5 y + 7)

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3 y (10 y - 7)

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3 y (7 y - 5)

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Solution

The correct option is D $3 y (5 y - 7)$
$39 y^{3} (50 y^{2} - 98) \div 26 y^{2} (5 y + 7)$
$= \frac{39 y^{3} (50 y^{2} - 98)}{26 y^{2} (5 y + 7)}$
$= \frac{39 y^{3} (5 \times 5 \times 2 y^{2} - 7 \times 7 \times 2)}{26 y^{2} (5 y + 7)}$
$= \frac{39 y^{3} \times 2 (5 y)^{2} - (7)^{2}}{26 y^{2} (5 y + 7)}$
$= \frac{39 y^{3} \times 2 (5 y + 7) (5 y - 7)}{26 y^{2} (5 y + 7)}$
$= 3 y (5 y - 7)$

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Similar questions

39 y^{3} (50 y^{2} - 98) \div 26 y^{2} (5 y + 7)

Factorise and divide the expressions given

i.) $(m^{2} - 11 m - 42) \div (m + 3)$

ii.) $39 y^{3} (50 y^{2} - 98) \div 26 y^{2} (5 y + 7)$ [4 MARKS]

Factorise the expressions and divide them as directed.

(i) (y² + 7y + 10) ÷ (y + 5)

(ii) (m² − 14m − 32) ÷ (m + 2)

(iii) (5p² − 25p + 20) ÷ (p − 1)

(iv) 4yz(z² + 6z − 16) ÷ 2y(z + 8)

(v) 5pq(p² − q²) ÷ 2p(p + q)

(vi) 12xy(9x² − 16y²) ÷ 4xy(3x + 4y)

(vii) 39y³(50y²− 98) ÷ 26y²(5y+ 7)

39 y^{3} (50 y^{2} - 98) \div 26 y^{2} (5 y + 7)

39 y^{3} (50 y^{2} - 98) \div 26 y^{2} (5 y + 7)

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