Question

$\Large{\underline{\underline{\bf{Question:-}}}}$
Factorise the following expressions:-
1) p² - 3pq + 2q²
2) x² + 11x + 30

★No Cópy !​

Answer 1

Answer:

ax2+bx (ii) 7p2+21q2 (iii) 2x3+2x y2+2xz2 (iv) am2+bm2+bn2+an2 (v) (lm+l)+m+ 1 (vi) y(y+z)+9(y+z) (vii) 5y2-20y-8z+2yz (viii) ..

Answer 2

p² - 3pq + 2q²

Consider p² - 3pq + 2q² as a polynomial over variable p.

$\sf \: {p}^{2} - 3qp \times 2 {q}^{2}$

p² -3qp + 2q²

Find one factor of the form p^k + m, where p^k divides the monomial with the highest power p² and m divides the constant factor 2q². One such factor is p-2q. Factor the polynomial by dividing it by this factor.

$(p - 2q)(p - q)$

x² + 11x + 30

Factor the expression by grouping. First, the expression needs to be rewritten as x² + ax + bx + 30. To find a and b, set up a system to be solved.

$\sf \: a + b = 11$

$\sf \: ab \: = 1 \times 30 = 30$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

$\sf1 \: , \: 30$

$2 \: , \: 15$

$3 \: ,\: 10$

$5 \: , \: 6$

Calculate the sum for each pair.

1 + 30 = 31

2 + 15 = 17

3 + 10 = 13

5 + 6 = 11

The solution is the pair that gives sum 11.

$a = 5$

$b = 6$

Rewrite x² + 11x + 30 as (x² + 5x) + (6x+30).

$\sf \: (x ^{2} + 5x) + (6x + 30)$

Factor out x in the first and 6 in the second group.

$x(x + 5) + 6(x + 5)$

Factor out common term x+5 by using distributive property.

$\sf(x + 5)(x + 6)$

$\color{lime} \boxed{hope \: it \: helps \: uhh : )}$