Factorise 6a−15a2+20ab−8b25a3−4a\dfrac{6a-15a^2+20ab-8b}{25a^3-4a} completely.

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Worked solution

4

1. Factorise the numerator by grouping.

6a−15a2+20ab−8b=3a(2−5a)+4b(5a−2)=3a(2−5a)−4b(2−5a)=(2−5a)(3a−4b)\begin{aligned}6a-15a^2+20ab-8b&=3a(2-5a)+4b(5a-2)\\&=3a(2-5a)-4b(2-5a)\\&=(2-5a)(3a-4b)\end{aligned}

2. Factorise the denominator completely.

25a3−4a=a(25a2−4)=a((5a)2−22)=a(5a−2)(5a+2)\begin{aligned}25a^3-4a&=a(25a^2-4)\\&=a\left((5a)^2-2^2\right)\\&=a(5a-2)(5a+2)\end{aligned}

3. Cancel the common factor and write the expression in simplest factorised form.

6a−15a2+20ab−8b25a3−4a=(2−5a)(3a−4b)a(5a−2)(5a+2)=−(5a−2)(3a−4b)a(5a−2)(5a+2)=−(3a−4b)a(5a+2)=4b−3aa(5a+2)\begin{aligned}\frac{6a-15a^2+20ab-8b}{25a^3-4a}&=\frac{(2-5a)(3a-4b)}{a(5a-2)(5a+2)}\\&=\frac{-(5a-2)(3a-4b)}{a(5a-2)(5a+2)}\\&=\frac{-(3a-4b)}{a(5a+2)}\\&=\frac{4b-3a}{a(5a+2)}\end{aligned}

Answer

4b−3aa(5a+2)\frac{4b-3a}{a(5a+2)}

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