How do you factor $ 15{x^2} - x - 2 $ by grouping?
Answer
579.6k+ views
Hint: First we will reduce the equation further if possible. Then we will try to factorise the terms in the equation. Split the middle term and factorise the equation. Then equate the factors equal to zero and evaluate the value of the variable.
Complete step-by-step answer:
We will start off by reducing any reducible terms in the equation.
$ 15{x^2} - x - 2 $
Now we will split the coefficient of the middle term in two parts. bvv vn hgvg jvh
$
15{x^2} - x - 2 \\
15{x^2} - 6x + 5x - 2 \;
$
Now we will group the like terms.
$
15{x^2} - 6x + 5x - 2 \\
3x(5x - 2) + (5x - 2) \\
(3x + 1)(5x - 2) \;
$
Now we will equate the factors with zero.
$
(3x + 1)(2x + 5) = 0 \\
x = - \dfrac{1}{3}, - \dfrac{5}{2} \;
$
Hence, the values of $ x $ are $ - \dfrac{1}{3}, - \dfrac{5}{2} $ .
So, the correct answer is “ $ - \dfrac{1}{3}, - \dfrac{5}{2} $ ”.
Note: Factorisation consists of writing a number or another mathematical objects as a product of several objects of the same kind. In particular, a univariate polynomial with complex coefficients admits a unique factorisation into linear polynomials; this is a version of the fundamental theorem of algebra.by the fundamental theorem of arithmetic, every integer greater than $ 1 $ has unique factorisation into prime numbers, which are those integers which cannot be further factored into the product of integers greater than one.
Complete step-by-step answer:
We will start off by reducing any reducible terms in the equation.
$ 15{x^2} - x - 2 $
Now we will split the coefficient of the middle term in two parts. bvv vn hgvg jvh
$
15{x^2} - x - 2 \\
15{x^2} - 6x + 5x - 2 \;
$
Now we will group the like terms.
$
15{x^2} - 6x + 5x - 2 \\
3x(5x - 2) + (5x - 2) \\
(3x + 1)(5x - 2) \;
$
Now we will equate the factors with zero.
$
(3x + 1)(2x + 5) = 0 \\
x = - \dfrac{1}{3}, - \dfrac{5}{2} \;
$
Hence, the values of $ x $ are $ - \dfrac{1}{3}, - \dfrac{5}{2} $ .
So, the correct answer is “ $ - \dfrac{1}{3}, - \dfrac{5}{2} $ ”.
Note: Factorisation consists of writing a number or another mathematical objects as a product of several objects of the same kind. In particular, a univariate polynomial with complex coefficients admits a unique factorisation into linear polynomials; this is a version of the fundamental theorem of algebra.by the fundamental theorem of arithmetic, every integer greater than $ 1 $ has unique factorisation into prime numbers, which are those integers which cannot be further factored into the product of integers greater than one.
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