Question

3m(3m - 1) 24(3m + 1)
(3m - 1)(3m + 4)

1. $5n^2 - 26n + 24$
2. $(21x^3 - 6x^2)(-7x + 2)$
3. $(21k^3 - 3k^2)(49k - 7)$

Explanation:

Problem 6: Factorize \(5n^2 - 26n + 24\)

Step 1: Multiply \(a\) and \(c\)

For quadratic \(ax^2+bx+c\), here \(a = 5\), \(b=-26\), \(c = 24\). Multiply \(a\times c=5\times24 = 120\).

Step 2: Find two numbers

Find two numbers that multiply to \(120\) and add to \(b=-26\). The numbers are \(-20\) and \(-6\) (since \(-20\times -6 = 120\) and \(-20 + (-6)=-26\)).

Step 3: Split the middle term

Rewrite the quadratic as \(5n^2-20n - 6n + 24\).

Step 4: Group and factor

Group the first two and last two terms: \((5n^2 - 20n)+(-6n + 24)\). Factor out common terms: \(5n(n - 4)-6(n - 4)\). Then factor out \((n - 4)\): \((5n - 6)(n - 4)\).

Problem 8: Factorize \((21x^3 - 6x^2)(-7x + 2)\)

Step 1: Factor out from first polynomial

Factor out \(3x^2\) from \(21x^3-6x^2\): \(3x^2(7x - 2)\).

Step 2: Rewrite second polynomial

Rewrite \(-7x + 2\) as \(-(7x - 2)\).

Step 3: Multiply the two polynomials

Now we have \(3x^2(7x - 2)\times(-(7x - 2))=-3x^2(7x - 2)^2\).

Problem 10: Factorize \((21k^3 - 3k^2)(49k - 7)\)

Step 1: Factor out from first polynomial

Factor out \(3k^2\) from \(21k^3-3k^2\): \(3k^2(7k - 1)\).

Step 2: Factor out from second polynomial

Factor out \(7\) from \(49k - 7\): \(7(7k - 1)\).

Step 3: Multiply the two polynomials

Multiply them: \(3k^2(7k - 1)\times7(7k - 1)=21k^2(7k - 1)^2\).

Answer:

s:

1. \(\boldsymbol{(5n - 6)(n - 4)}\)
2. \(\boldsymbol{-3x^2(7x - 2)^2}\)
3. \(\boldsymbol{21k^2(7k - 1)^2}\)