Question

Factorization : 3a2-10a+8

Answer 1

Explanation:

3a² - 10a + 8

1. find delta and √delta

∆=(-10)²-4*3*8=100-96=4

√∆=2

2. find a1, a2 with formulas (because ∆>0)

a1= (10+2)/6 = 2

a2=(10-2)/6=4/3

Note: ka²+la+m = k(a-a1)(a-a2)

So answer is 3(a-2)(a-4/3)=(3a-4)(a-2).

OR: we can find a root from factors 8 (it is 2) and then:

3a²-10a+8= 3a² - 6a -4a + 8 = 3a(a-2)-4(a-2)=(3a-4)(a-2)

Answer 2

(a - 2) • (3a - 4)

Explanation:

(3a2 - 10a) + 8

Trying to factor by splitting the middle term

2.1 Factoring 3a2-10a+8

The first term is, 3a2 its coefficient is 3 .

The middle term is, -10a its coefficient is -10 .

The last term, "the constant", is +8

Multiply the coefficient of the first term by the constant 3 • 8 = 24

Find two factors of 24 whose sum equals the coefficient of the middle term, which is -10 .

-24 + -1 = -25

-12 + -2 = -14

-8 + -3 = -11

-6 + -4 = -10 That's it

Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -6 and -4

3a2 - 6a - 4a - 8

Add up the first 2 terms, pulling out like factors :

3a • (a-2)

Add up the last 2 terms, pulling out common factors :

4 • (a-2)

Add up the four terms of step 4 :

(3a-4) • (a-2)

Which is the desired factorization

Final result :

(a - 2) • (3a - 4)