factorise 29+a-29abc by regrouping method
Answers
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:
29+43÷33=12378
Step-by-step explanation:
first mutlsh then dibede