Sure, let's factorize each of the given quadratics by splitting the middle term.
### (a) [tex]\( 3x^2 + 11x + 30 \)[/tex]
1. To factorize the quadratic [tex]\( 3x^2 + 11x + 30 \)[/tex], first look for two numbers that multiply to [tex]\( 3 \times 30 = 90 \)[/tex] and add up to 11. However, upon inspection, it is seen that there are no such pairs of integers. Therefore, this quadratic expression cannot be factorized over the integers.
So, the factorization does not simplify further in this case.
### (b) [tex]\( 2\sqrt{2}x^2 + 9x + 5\sqrt{2} \)[/tex]
1. Similarly, consider the quadratic [tex]\( 2\sqrt{2}x^2 + 9x + 5\sqrt{2} \)[/tex].
2. Look for two numbers that multiply to [tex]\( 2\sqrt{2} \times 5\sqrt{2} = 20 \)[/tex] and add up to [tex]\( 9 \)[/tex]. Again, verifying all possibilities shows no such pairs of integers exist. Hence, this quadratic also cannot be factorized over the integers.
So, the factorization does not simplify further for this quadratic polynomial.
### (c) [tex]\( 4x^2 - 13x + 10 \)[/tex]
1. Look for two numbers that multiply to [tex]\( 4 \times 10 = 40 \)[/tex] and add up to [tex]\( -13 \)[/tex]. These numbers are [tex]\( -5 \)[/tex] and [tex]\( -8 \)[/tex].
2. Rewrite the middle term as the sum of these two numbers:
[tex]\[ 4x^2 - 8x - 5x + 10 \][/tex]
3. Factor by grouping:
[tex]\[ 4x(x - 2) - 5(x - 2) \][/tex]
4. Factor out the common binomial factor:
[tex]\[ (4x - 5)(x - 2) \][/tex]
So, the factorization of [tex]\( 4x^2 - 13x + 10 \)[/tex] is:
[tex]\[ (4x - 5)(x - 2) \][/tex]
