Question

Factorise using factor theorem
9a^3-19a+10​

Answer 1

Answer:

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Step-by-step explanation:

9 a ^ { 3 } - 19 a + 10

9a

3

−19a+10

\frac{p}{q}

p10q9-\frac{5}{3}\approx -1.666666667

3a+5

\left(3a+5\right)\left(3a^{2}-5a+2\right)

(3a+5)(3a

2

−5a+2)

3a^{2}-5a+2

3a^{2}+pa+qa+2

pq

p+q=-5

p+q=−5

pq=3\times 2=6

pq=3×2=6

-1,-6

−1,−6

-2,-3

−2,−3

-1-6=-7

−1−6=−7

-2-3=-5

−2−3=−5

-5

p=-3

p=−3

q=-2

q=−2

3a^{2}-5a+2

\left(3a^{2}-3a\right)+\left(-2a+2\right)

\left(3a^{2}-3a\right)+\left(-2a+2\right)

(3a

2

−3a)+(−2a+2)

3a-2

3a\left(a-1\right)-2\left(a-1\right)

3a(a−1)−2(a−1)

a-1

\left(a-1\right)\left(3a-2\right)

(3a−2)(a−1)(3a+5)