Question

Factorise: 3a^2b^2+17ab-56​

Answer 1

Following are the steps for getting the answer:

Given:

$3a^2b^2+17ab-56$

to find:

factorize $3a^2b^2+17ab-56$

Solution:

we have to find the factor of $3a^2b^2+17ab-56$

so,

$3a^2b^2+17ab-56$

we can write it like

3a²b²+24ab-7ab-56

(∴24ab-7ab=17ab)

Now will use the grouping method

=(3a²b²+24ab)(-7ab+56)

=3ab(ab+8)-7(ab+8)

=(3ab-7)(ab+8)

thus, the factor of $3a^2b^2+17ab-56$  is (3ab-7)(ab+8)

Answer 2

Answer:

The factors are $(ab+8),(3ab-7)$.

Step-by-step explanation:

Given equation is $3a^{2}b^2+17ab-56$.

We need to factorize the given equation.

Factorization or factoring consists of writing a number or another mathematical object as a product of several factors.

While factoring a polynomial,

we have to find two numbers whose product is equal to the product of the coefficients of first and last term. Also, the sum of two numbers should be equal to the coefficient of the middle term.

Let the two numbers be $x,y$.

$x+y=17\\xy=-168$

The factors of 168 are

$1,2,3,4,6,7,8,12,14,21,24,28,42,56,84,168$

Of all the factors of 168 only 24 and -7 fit the above scenario.

The given equation becomes,

$=3a^2b^2+24ab-7ab-56\\=3ab(ab+8)-7(ab+8)\\=(ab+8)(3ab-7)$

The final answer is $3a^{2}b^2+17ab-56=(ab+8)(3ab-7)$.