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To factorize the quadratic polynomial [tex]\( 10x^2 - 9x + 2 \)[/tex], let's follow these steps:
1. Understand the given polynomial: We have the polynomial in the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 10 \)[/tex], [tex]\( b = -9 \)[/tex], and [tex]\( c = 2 \)[/tex].
2. Find the roots of the polynomial: The polynomial [tex]\( 10x^2 - 9x + 2 \)[/tex] can be factored using the roots of the equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
The quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] provides the roots of the polynomial.
3. Calculate the discriminant: For [tex]\( 10x^2 - 9x + 2 \)[/tex]:
[tex]\[ \text{Discriminant} = b^2 - 4ac = (-9)^2 - 4(10)(2) = 81 - 80 = 1. \][/tex]
4. Calculate the roots using the quadratic formula:
[tex]\[ x = \frac{-(-9) \pm \sqrt{1}}{2 \cdot 10} = \frac{9 \pm 1}{20}. \][/tex]
This gives us the roots:
[tex]\[ x = \frac{9 + 1}{20} = \frac{10}{20} = \frac{1}{2} \][/tex]
and
[tex]\[ x = \frac{9 - 1}{20} = \frac{8}{20} = \frac{4}{10} = \frac{2}{5}. \][/tex]
5. Express the polynomial in its factorized form: The factors are found using the roots [tex]\( x = \frac{1}{2} \)[/tex] and [tex]\( x = \frac{2}{5} \)[/tex].
Thus, the polynomial can be written as:
[tex]\[ 10x^2 - 9x + 2 = a (x - r_1)(x - r_2), \][/tex]
where [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] are the roots. Substituting [tex]\( r_1 = \frac{1}{2} \)[/tex] and [tex]\( r_2 = \frac{2}{5} \)[/tex], and noting [tex]\( a = 10 \)[/tex]:
[tex]\[ 10(x - \frac{1}{2})(x - \frac{2}{5}). \][/tex]
6. Simplify the factors: Multiple linear fractions need to be transformed back into polynomial factors:
[tex]\[ 10\left(x - \frac{1}{2}\right)\left(x - \frac{2}{5}\right) = 10\left(\frac{2x - 1}{2}\right)\left(\frac{5x - 2}{5}\right). \][/tex]
When simplified, the factors become:
[tex]\[ = (2x - 1)(5x - 2). \][/tex]
So, the polynomial [tex]\( 10x^2 - 9x + 2 \)[/tex] can be factorized as:
[tex]\[ 10x^2 - 9x + 2 = (2x - 1)(5x - 2). \][/tex]
Therefore, the factorized form of the polynomial is:
[tex]\[ \boxed{(2x - 1)(5x - 2)}. \][/tex]
1. Understand the given polynomial: We have the polynomial in the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 10 \)[/tex], [tex]\( b = -9 \)[/tex], and [tex]\( c = 2 \)[/tex].
2. Find the roots of the polynomial: The polynomial [tex]\( 10x^2 - 9x + 2 \)[/tex] can be factored using the roots of the equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
The quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] provides the roots of the polynomial.
3. Calculate the discriminant: For [tex]\( 10x^2 - 9x + 2 \)[/tex]:
[tex]\[ \text{Discriminant} = b^2 - 4ac = (-9)^2 - 4(10)(2) = 81 - 80 = 1. \][/tex]
4. Calculate the roots using the quadratic formula:
[tex]\[ x = \frac{-(-9) \pm \sqrt{1}}{2 \cdot 10} = \frac{9 \pm 1}{20}. \][/tex]
This gives us the roots:
[tex]\[ x = \frac{9 + 1}{20} = \frac{10}{20} = \frac{1}{2} \][/tex]
and
[tex]\[ x = \frac{9 - 1}{20} = \frac{8}{20} = \frac{4}{10} = \frac{2}{5}. \][/tex]
5. Express the polynomial in its factorized form: The factors are found using the roots [tex]\( x = \frac{1}{2} \)[/tex] and [tex]\( x = \frac{2}{5} \)[/tex].
Thus, the polynomial can be written as:
[tex]\[ 10x^2 - 9x + 2 = a (x - r_1)(x - r_2), \][/tex]
where [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] are the roots. Substituting [tex]\( r_1 = \frac{1}{2} \)[/tex] and [tex]\( r_2 = \frac{2}{5} \)[/tex], and noting [tex]\( a = 10 \)[/tex]:
[tex]\[ 10(x - \frac{1}{2})(x - \frac{2}{5}). \][/tex]
6. Simplify the factors: Multiple linear fractions need to be transformed back into polynomial factors:
[tex]\[ 10\left(x - \frac{1}{2}\right)\left(x - \frac{2}{5}\right) = 10\left(\frac{2x - 1}{2}\right)\left(\frac{5x - 2}{5}\right). \][/tex]
When simplified, the factors become:
[tex]\[ = (2x - 1)(5x - 2). \][/tex]
So, the polynomial [tex]\( 10x^2 - 9x + 2 \)[/tex] can be factorized as:
[tex]\[ 10x^2 - 9x + 2 = (2x - 1)(5x - 2). \][/tex]
Therefore, the factorized form of the polynomial is:
[tex]\[ \boxed{(2x - 1)(5x - 2)}. \][/tex]
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