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Sagot :
To factor the quadratic expression [tex]\( 5x^2 + 27x + 10 \)[/tex], follow these steps:
1. Identify the coefficients:
The quadratic expression is in the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = 27 \)[/tex], and [tex]\( c = 10 \)[/tex].
2. Set up the factorization format:
To factor [tex]\( 5x^2 + 27x + 10 \)[/tex], we need to write it in the form [tex]\( (mx + p)(nx + q) \)[/tex], where [tex]\( m \)[/tex], [tex]\( n \)[/tex], [tex]\( p \)[/tex], and [tex]\( q \)[/tex] are constants to be determined.
3. Find the pair of binomials:
We need to find two numbers that multiply to [tex]\( a \cdot c \)[/tex] (the product of the first and last coefficients) and add up to [tex]\( b \)[/tex] (the middle coefficient).
Here, [tex]\( a \cdot c = 5 \times 10 = 50 \)[/tex] and [tex]\( b = 27 \)[/tex].
4. Determine the pairs that multiply to 50 and add up to 27:
After checking possible pairs, we find that the pair (2, 25) works because:
[tex]\[ 2 \times 25 = 50 \quad \text{and} \quad 2 + 25 = 27. \][/tex]
5. Rewrite the middle term:
Use the pair (2, 25) to split the middle term [tex]\( 27x \)[/tex]:
[tex]\[ 5x^2 + 27x + 10 = 5x^2 + 2x + 25x + 10. \][/tex]
6. Factor by grouping:
Group the terms to factor by grouping:
[tex]\[ = (5x^2 + 2x) + (25x + 10). \][/tex]
Factor out the common terms in each group:
[tex]\[ = x(5x + 2) + 5(5x + 2). \][/tex]
7. Factor out the common binomial factor:
Both groups contain the common binomial factor [tex]\( (5x + 2) \)[/tex]:
[tex]\[ = (x + 5)(5x + 2). \][/tex]
So, the quadratic expression [tex]\( 5x^2 + 27x + 10 \)[/tex] factors to [tex]\( (x + 5)(5x + 2) \)[/tex].
Hence, the factored form is:
[tex]\[ (x + 5)(5x + 2) \][/tex]
1. Identify the coefficients:
The quadratic expression is in the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = 27 \)[/tex], and [tex]\( c = 10 \)[/tex].
2. Set up the factorization format:
To factor [tex]\( 5x^2 + 27x + 10 \)[/tex], we need to write it in the form [tex]\( (mx + p)(nx + q) \)[/tex], where [tex]\( m \)[/tex], [tex]\( n \)[/tex], [tex]\( p \)[/tex], and [tex]\( q \)[/tex] are constants to be determined.
3. Find the pair of binomials:
We need to find two numbers that multiply to [tex]\( a \cdot c \)[/tex] (the product of the first and last coefficients) and add up to [tex]\( b \)[/tex] (the middle coefficient).
Here, [tex]\( a \cdot c = 5 \times 10 = 50 \)[/tex] and [tex]\( b = 27 \)[/tex].
4. Determine the pairs that multiply to 50 and add up to 27:
After checking possible pairs, we find that the pair (2, 25) works because:
[tex]\[ 2 \times 25 = 50 \quad \text{and} \quad 2 + 25 = 27. \][/tex]
5. Rewrite the middle term:
Use the pair (2, 25) to split the middle term [tex]\( 27x \)[/tex]:
[tex]\[ 5x^2 + 27x + 10 = 5x^2 + 2x + 25x + 10. \][/tex]
6. Factor by grouping:
Group the terms to factor by grouping:
[tex]\[ = (5x^2 + 2x) + (25x + 10). \][/tex]
Factor out the common terms in each group:
[tex]\[ = x(5x + 2) + 5(5x + 2). \][/tex]
7. Factor out the common binomial factor:
Both groups contain the common binomial factor [tex]\( (5x + 2) \)[/tex]:
[tex]\[ = (x + 5)(5x + 2). \][/tex]
So, the quadratic expression [tex]\( 5x^2 + 27x + 10 \)[/tex] factors to [tex]\( (x + 5)(5x + 2) \)[/tex].
Hence, the factored form is:
[tex]\[ (x + 5)(5x + 2) \][/tex]
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