Question

factorise 29+a-29abc by regrouping method

Answer 1

Answer:

(i) a2 + bc + ab + ac

Solution:

The expression is a2 + bc + ab + ac

By suitably rearranging the terms, we have;

= a2 + ab + ac + bc

= a(a + b) + c(a + b)

= (a + b) (a + c).

(ii) ax2 + by2 + bx2 + ay2

Solution:

The expression is ax2 + by2 + bx2 + ay2

By suitably rearranging the terms, we have;

= ax2 + ay2 + bx2 + by2

= a(x2 + y2) + b(x2 + y2)

= (x2 + y2) (a + b).

2. Factor grouping the algebraic expressions:

(i) xy - pq + qy - px

Solution:

xy - pq + qy - px

By suitably rearranging the terms, we have;

= (xy - px) + (qy - pq)

= x (y - p) + q (y - p)

= (y - p) (x + q).

Therefore by factoring expressions we get (y - p) (x + q).

(ii) ab(x2 + y2) + xy(a2 + b2).

Solution:

ab(x2 + y2) + xy(a2 + b2)

By suitably rearranging the terms, we have;

= abx2 + aby2 + a2xy + b2xy

= (abx2 + a2xy) + (aby2 + bxy)

= ax(bx + ay) + by(ay + bx)

= ax(bx + ay) + by(bx + ay)

= (bx + ay) (ax + by).

Therefore by factoring expressions we get (bx + ay) (ax + by)

Answer 2

Answer:

29+43÷33=12378

Step-by-step explanation:

first mutlsh then dibede