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Sagot :
Answer:
By finding the two values that when multiplied and added gives the value for (a * c) when multiplied and b when added
Step-by-step explanation:
Given:
[tex]110x^2 + 192x + 72 \\\\ \left[\begin{array}{c}a=110x^2\\b=192x\\c=72\end{array}\right][/tex]
To factorize it, we need to know what values we will multiply and the result gives [tex]72 \times 110x^2= 7920x^2[/tex] and add to give [tex]192x[/tex]
After Trial and Error the values are 132x, 60x
[tex]110x^2 + 132x + 60x + 72[/tex]
Simplifying
[tex]11x(10x + 12) + 6(10x + 12)[/tex]
[tex](11x + 6)(10x + 12)[/tex]
Therefore, the end value matches the answer given verifying our calculations.
Answer:
(11x + 6)(10x + 12)
Step-by-step explanation:
To factorise the quadratic expression 110x² + 192x + 72, we need to find two binomials whose product equals this expression.
As all the coefficients of 110x² + 192x + 72 are divisible by 2, we can begin by factoring out the greatest common factor (GCF), which is 2:
[tex]2(55x^2+96x+36)[/tex]
Now, we can focus on factoring the trinomial inside the parentheses.
To factorise a quadratic expression in the form ax² + bx + c, we need to find two numbers that multiply to the product of a and c, and sum to b.
In the case of 55x² + 96x + 36, the coefficients are:
- a = 55
- b = 96
- c = 36
Therefore:
[tex]a \times c = 55 \times 36= 1980\\\\b = 96[/tex]
So, we are looking for two numbers that multiply to 1980 and sum to 96.
The factors of 1980 are:
- 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 18, 20, 22, 30, 33, 36, 44, 45, 55, 60, 66, 90, 99, 110, 132, 165, 180, 198, 220, 330, 396, 495, 660, 990, 1980.
Therefore, the factor pair that sums to 96 is:
- 30 and 66
So, we can rewrite 96x in the trinomial as 30x + 66x:
[tex]2(55x^2+30x+66x+36)[/tex]
Now, factor the first two terms and the last two terms separately:
[tex]2(5x(11x+6)+6(11x+6))[/tex]
Factor out the common term (11x + 6):
[tex]2(11x+6)(5x+6)[/tex]
Multiply the binomial (5x + 6) by 2:
[tex](11x+6)(10x+12)[/tex]
Therefore, the factorisation of 110x² + 192x + 72 is:
[tex]\Large\boxed{\boxed{(11x+6)(10x+12)}}[/tex]
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